Gas formation from reaction #1

The reaction between copper(II)sulfide and nitric acid gives copper ion, sulfur, nitrogen monoxide and water. If 71.5gr of copper(II)sulfide react with 100gr of nitric acid, compute the mass of sulfur and the volume of nitrogen monoxide that form.

SOLUTION:
The reaction is:
$$CuS+HNO_3 \rightarrow Cu^{++}+S+NO+H_2O$$
You can easily balance it:
$$\underset{100gr}{\underbrace{CuS}}+\underset{71.5gr}{\underbrace{{\color{red} 2}H^++{\color{red} 2}NO_3^-}} \rightarrow Cu^{++}+\underset{m?}{\underbrace{S}}+\underset{V?}{\underbrace{{\color{red} 2}NO}}+H_2O$$

We can calculate the number of mol (n) of CuS and $HNO_3$:

$$n_{CuS}=\frac{m_{CuS}}{MW_{CuS}}=\frac{71.5gr}{95.62gr/mol}=0.748mol$$

$$n_{HNO_3}=\frac{m_{HNO_3}}{MW_{HNO_3}}=\frac{100gr}{63gr/mol}=1.587mol$$

For every mol of CuS we obtain one mol of sulfur: $n_{CuS}=n_{S}$. So the mass of sulfur is:
$$m_S=n_S \cdot MW_S=0.748mol \cdot 32.07gr/mol=23.99gr$$

The NO forms from nitric acid. The number of moles of nitric acid is equal to the number of moles of NO: $n_{HNO_3}=n_{NO}$. The volume of one mole of gas is equal to 22.4 liters at standard conditions (0°C, 1atm):
$$V_{NO}=n_{NO} \cdot 22.4L/mol=1.587mol\cdot22.4L/mol=34.55L$$

---

PROBLEM 18:
2.0 grams of propane react with 7.0 grams of oxygen; suppose that the reaction is complete, and forms water and CO. Calculate the mass of water and the volume of CO and $O_2$.
Continue reading »

Redox equation and ionic form #4

Balance this equation and put it in the ionic form:
$$MnI_2+NaBiO_3+H_2SO_4 \rightarrow$$
$$\rightarrow \quad\quad NaMnO_4+NaIO_3+Bi_2(SO_4)_3+Na_2SO_4+H_2O$$

SOLUTION:
Using the inspection technique we have:
$${\color{red} a}MnI_2+{\color{red} b}NaBiO_3+{\color{red} c}H_2SO_4 \rightarrow$$
$$\rightarrow \quad\quad {\color{red} d}NaMnO_4+{\color{red} e}NaIO_3+{\color{red} f}Bi_2(SO_4)_3+{\color{red} g}Na_2SO_4+{\color{red} h}H_2O$$
$$\begin{cases}Mn:a=d\\ I:2a=e\\ Na:b=d+e+2g\\ Bi:b=2f\\ S:c=3f+g\\ H:2c=2h\\ O:3b+4c=4d+3e+12f+4g+h \end{cases}$$

This is a reduction-oxidation equation, in fact:
$$\overset{+2}{Mn}\overset{-1}{I_2}+Na\overset{+5}{Bi}O_3+H_2SO_4 \rightarrow$$
$$\rightarrow \quad\quad Na\overset{+7}{Mn}O_4+Na\overset{+5}{I}O_3+\overset{+3}{Bi_2}(SO_4)_3+Na_2SO_4+H_2O$$

- Manganese changes its oxidation number from +2 to +7: this is the first oxidation (Manganese is the reductant).
- Iodine changes its oxidation number from -1 to +5: this is the second oxidation (Iodine is the reductant).
- Bismuth changes its oxidation number from +5 to +3: this is the reduction (Bismuth is the oxidant).

Write down the three half-equations:
$$\begin{cases} \overset{+2}{Mn} \rightarrow \overset{+7}{Mn}+5e^-\\\\ \overset{-1}{I_2} \rightarrow 2\overset{+5}{I}+12e^-\\\\ 2\overset{+5}{Bi}+4e^- \rightarrow \overset{+3}{Bi_2}\end{cases}$$

Now sum the two oxidation equation:
$$\begin{cases} \overset{+2}{Mn}+\overset{-1}{I_2} \rightarrow \overset{+7}{Mn}+2\overset{+5}{I}+17e^-\\\\ 2\overset{+5}{Bi}+4e^- \rightarrow \overset{+3}{Bi_2}\end{cases}$$

Now balance the electrons and sum:
$$\begin{cases} \overset{+2}{Mn}+\overset{-1}{I_2} \rightarrow \overset{+7}{Mn}+2\overset{+5}{I}+17e^- & \quad \mbox{(x4)}\\\\ 2\overset{+5}{Bi}+4e^- \rightarrow \overset{+3}{Bi_2} & \quad \mbox{(x17)}\end{cases}+$$
$$\rule[0.3cm]{8cm}{0.01cm} $$
$$= \quad\quad 4\overset{+2}{Mn}\overset{-1}{I_2}+34\overset{+5}{Bi} \rightarrow 4\overset{+7}{Mn}+8\overset{+5}{I}+17\overset{+3}{Bi_2}$$

So we have calculated the following stoichiometric coefficients:
$$\begin{cases}a=4\\ b=34\\ d=4\\ e=8\\ f=17 \end{cases}$$
The other stoichiometric coefficients are:
$$\begin{cases}g=11\\ c=62\\ h=62 \end{cases}$$

The balanced equation is:
$${\color{red} 4}MnI_2+{\color{red} 34}NaBiO_3+{\color{red} 62}H_2SO_4 \rightarrow$$
$$\rightarrow \quad\quad {\color{red} 4}NaMnO_4+{\color{red} 8}NaIO_3+{\color{red} 17}Bi_2(SO_4)_3+{\color{red} 11}Na_2SO_4+{\color{red} 62}H_2O$$

Now we write it in the ionic form:
$$ 4MnI_2++34Na^++34BiO_3^-+124H^++62SO_4^{2-} \rightarrow$$
$$\rightarrow \quad\quad 4Na^++4MnO_4^-+8Na^++8IO_3^-+34Bi^{3+}+$$
$$+ \quad\quad\quad\quad 51SO_4^{2-}+22Na^++11SO_4^{2-}+62H_2O$$
Now sum the sodium and the solfate ions and simplify:
$$4MnI_2+34BiO_3^-+124H^+ \rightarrow$$
$$\rightarrow \quad 4MnO_4^-+8IO_3^-+34Bi^{3+}+62H_2O$$
We can divide by two all the stoichiometric coefficients. The net ionic equation balanced is:
$$2MnI_2+17BiO_3^-+62H^+ \rightarrow$$
$$\rightarrow \quad 2MnO_4^-+4IO_3^-+17Bi^{3+}+31H_2O$$

---

PROBLEM 17:
Balance the following redox equation and put it in the ionic form:
$$[Ni(NH_3)_4]Cl_2+(C_6H_5)NC+H_2O \rightarrow $$
$$\quad Ni[(C_6H_5)NC]_4+NH_3+NH_4Cl+(C_6H_5)NCO$$
Continue reading »

Redox equation and ionic form #3

Balance the following equation and put it in the ionic form:
$$Zn+NaNO_3+NaOH+H_2O \rightarrow Na_2[Zn(OH)_4]+NH_3$$

SOLUTION:
First we write down the relationships between the stoichiometric coefficients for the balance of mass:
$${\color{red} a}Zn+{\color{red} b}NaNO_3{\color{red} c}+NaOH+{\color{red} d}H_2O \rightarrow {\color{red} e}Na_2[Zn(OH)_4]+{\color{red} f}NH_3$$
$$\begin{cases} Zn:a=e\\ Na:b+c=2e\\ N:b=f\\ H:c+2d=4e+3f\\ O:3b+c+d=4e\end{cases}$$

To verify if this is a redox equation, write down all oxidation number:
$$\overset{0}{Zn}+Na\overset{+5}{N}O_3+NaOH+H_2O \rightarrow Na_2[\overset{+2}{Zn}(OH)_4]+\overset{-3}{N}H_3$$

- Zinc changes its oxidation number from 0 to +2: this is the oxidation (Zinc is the reductant).
- Nitrogen changes its oxidation number from +5 to -3: this is the reduction (Nitrogen is the oxidant).

Now write down the two half-equations:
$$\begin{cases} \overset{0}{Zn} \rightarrow \overset{+2}{Zn}+2e^-\\\\ \overset{+5}{N}+8e^- \rightarrow \overset{-3}{N}\end{cases}$$

Balance the electrons and sum:
$$\begin{cases} \overset{0}{Zn} \rightarrow \overset{+2}{Zn}+2e^- & \quad \mbox{(x4)}\\\\ \overset{+5}{N}+8e^- \rightarrow \overset{-3}{N}\end{cases} \quad+$$
$$\rule[0.3cm]{6.5cm}{0.01cm} $$
$$=\quad\quad 4\overset{0}{Zn}+\overset{+5}{N} \rightarrow 4\overset{+2}{Zn}+\overset{-3}{N}$$

The stoichiometric coefficients found are:
$$\begin{cases}a=4\\ b=1\\e=4\\f=1 \end{cases}$$

From the relationships previously take, we have:
$$\begin{cases}Na:b+c=2e\\ O:3b+c+d=4e \end{cases} \Rightarrow \begin{cases}c=2e-b=7\\ d=4e-3b-c=6\end{cases}$$

The balanced redox equation is:
$${\color{red} 4}Zn+NaNO_3{\color{red} 7}+NaOH+{\color{red} 6}H_2O \rightarrow {\color{red} 4}Na_2[Zn(OH)_4]+NH_3$$

Now put it in the ionic form:
$$4Zn+Na^++NO_3^-+7Na^++7OH^-+6H_2O \rightarrow 8Na^++4[Zn(OH)_4]^{2-}+NH_3$$
Now simplify the sodium:
$$4Zn+NO_3^-+7OH^-+6H_2O \rightarrow 4[Zn(OH)_4]^{2-}+NH_3$$
Note that the charges are balanced.

---

PROBLEM 16:
Balance the following redox equation and put it in the ionic form:
$$As_2O_3+KIO_4 \rightarrow K_3AsO_4+KIO_2+H_2O$$
Continue reading »

Redox equation and ionic form #2

Balance the stoichiometric coefficients of this redox equation, and rewrite it in the ionic form:
$$Sn+HNO_3+H_2O \rightarrow H_2SnO_3+NO$$

SOLUTION:
For the principle of conservation of mass we have:
$${\color{red} a}Sn+{\color{red} b}HNO_3+{\color{red} c}H_2O \rightarrow {\color{red} d}H_2SnO_3+{\color{red} e}NO$$
$$\begin{cases} Sn:a=d\\ N:b=e\\ H:b+2c=2d\\ O:3b+c=3d+e\end{cases}$$

This is a redox equation, infact Sn and N change their oxidation number:
$$\overset{0}{Sn}+H\overset{+5}{N}O_3+H_2O \rightarrow H_2\overset{+4}{Sn}O_3+\overset{+2}{N}O$$

- Tin changes its oxidation number from 0 to +4: this is the oxidation (it is the reductant).
- Nitrogen changes its oxidation number from +5 to +2: this is the reduction (it is the oxidant).

So write down the two half-equations:
$$\begin{cases} \overset{0}{Sn} \rightarrow \overset{+4}{Sn}+4e^-\\\\ \overset{+5}{N}+3e^- \rightarrow \overset{+2}{N}\end{cases}$$

Now balance the electrons and sum:
$$\begin{cases} \overset{0}{Sn} \rightarrow \overset{+4}{Sn}+4e^- & \quad \mbox{(x3)}\\\\ \overset{+5}{N}+3e^- \rightarrow \overset{+2}{N} & \quad \mbox{(x4)}\end{cases} \quad+$$
$$\rule[0.3cm]{6.5cm}{0.01cm} $$
$$=\quad\quad 3\overset{0}{Sn}+4\overset{+5}{N} \rightarrow 3\overset{+4}{Sn}+4\overset{+2}{N}$$

From this equation we have found 4 stoichiometric coefficients:
$$\begin{cases} a=3\\ b=4\\ d=3\\ e=4\end{cases}$$
From the hydrogen mass relationship, we calculate $c=1$.

The balanced equation is:
$${\color{red} 3}Sn+{\color{red} 4}HNO_3+H_2O \rightarrow {\color{red} 3}H_2SnO_3+{\color{red} 4}NO$$

Now put it in the ionic form:
$$3Sn+4H^++4NO_3^-+2H_2O \rightarrow 6H^++3SnO_3^{2-}+4NO$$
We can simplify the $4H^++H_2O$ in reagents with the $6H^+$ in products. The net ionic equation is:
$$3Sn+4NO_3^- \rightarrow 3SnO_3^{2-}+4NO$$

---

PROBLEM 15:
Balance the following redox equation and put it in the ionic form, if possible:
$$Al+K_2SiF_6 \rightarrow AlF_3+Si+KF$$
Continue reading »

Redox equation and ionic form #1

Rewrite the following equation showing all substances in their ionic form in solution (the ionic equation) and balance it:
$$Hg(NO_3)_2+KI \rightarrow K_2HgI_4+KNO_3$$

SOLUTION:
The ionic equation is:
$$Hg^{2+}+2NO_3^-+K^++I^- \rightarrow 2K^++HgI_4^{2-}+K^++NO_3^-$$

Note that in reactants there are two $NO_3^-$ and only one $NO_3^-$ in products. So multiply by two $KNO_3$ in products:
$$Hg^{2+}+2NO_3^-+K^++I^- \rightarrow 2K^++HgI_4^{2-}+2K^++2NO_3^-$$

Now sum the same species -$K^+$- and simplify -$2NO_3^-$-:
$$Hg^{2+}+K^++I^- \rightarrow 4K^++HgI_4^{2-}$$

Now there are $4K^+$ in products. So multiply by four the $KI$ in reagents:
$$Hg^{2+}+4K^++4I^- \rightarrow 4K^++HgI_4^{2-}$$

Now simplify $K^+$:
$$Hg^{2+}+4I^- \rightarrow HgI_4^{2-}$$
This is the net ionic equation. Note that che charges are balanced.

---

PROBLEM 14:
Balance the following redox equation and put it in the ionic form:
$$Na+H_2O \rightarrow NaOH+H_2$$
Continue reading »

How to balance redox equations #4

Balance the following chemical equation:
$$CrI_3+NaNO_3+Na_2CO_3 \rightarrow Na_2CrO_4+NaIO_3+NaNO_2+CO_2$$

SOLUTION:
Using the inspection technique, we have:
$${\color{red} a}CrI_3+{\color{red} b}NaNO_3+{\color{red} c}Na_2CO_3 \rightarrow {\color{red} d}Na_2CrO_4+{\color{red} e}NaIO_3+{\color{red} f}NaNO_2+{\color{red} g}CO_2$$
$$\begin{cases} I:3a=e\\ Cr:a=d\\ Na:b+2c=2d+e+f\\ C:c=g\\ O:3b+3c=4d+3e+2f+2g\end{cases}$$
We have more variable than equation. So we can balance the chemical equation using the only inspection technique.

Is this a redox equation? Find out the oxidation number:

$$\overset{+3}{Cr}\overset{-1}{I_3}+Na\overset{+5}{N}O_3+Na_2CO_3 \rightarrow Na_2\overset{+6}{Cr}O_4+Na\overset{+5}{I}O_3+Na\overset{+3}{N}O_2+CO_2$$

Here there are one elements that is reducting (N), and TWO elements that are oxidating (Cr and I). Write down the three equations:

$$\begin{cases} \overset{+5}{N}+2e^- \rightarrow \overset{+3}{N}\\\\ \overset{+3}{Cr} \rightarrow \overset{+6}{Cr}+3e^-\\\\ \overset{-1}{I_3} \rightarrow 3\overset{+5}{I}+18e^-\end{cases}$$

Now simply sum the two oxidation reaction:

$$\begin{cases} \overset{+5}{N}+2e^- \rightarrow \overset{+3}{N}\\\\ \overset{+3}{Cr}+\overset{-1}{I_3} \rightarrow \overset{+6}{Cr}+3\overset{+5}{I}+21e^-\end{cases}$$

Now balance the electrons and sum the two half-equations:

$$\begin{cases} \overset{+5}{N}+2e^- \rightarrow \overset{+3}{N} & \quad \mbox{(x21)}\\\\ \overset{+3}{Cr}+\overset{-1}{I_3} \rightarrow \overset{+6}{Cr}+3\overset{+5}{I}+21e^- & \quad \mbox{(x2)}\end{cases}\quad+$$
$$\rule[0.3cm]{8cm}{0.01cm} $$
$$=\quad\quad 21\overset{+5}{N}+2\overset{+3}{Cr}+2\overset{-1}{I_3} \rightarrow 21\overset{+3}{N}+2\overset{+6}{Cr}+6\overset{+5}{I}$$

We now have:
$$\begin{cases} b=21\\ a=2\\ f=21\\ d=2\\ e=6\end{cases}$$

From the relationships previously found, we can obtain:

$$\begin{cases} c=5\\ g=5\end{cases}$$

The balanced equation is:
$${\color{red} 2}CrI_3+{\color{red} 21}NaNO_3+{\color{red} 5}Na_2CO_3 \rightarrow {\color{red} 2}Na_2CrO4+{\color{red} 6}NaIO3+{\color{red} 21}NaNO_2+{\color{red} 5}CO_2$$

---

PROBLEM 13:
Balance the following reaction:
sodium perchlorate reacts with ammonium chloride giving sodium chloride, nitrogen gas, hydrochloric acid and water.
Continue reading »

How to balance redox equations #3

Balance the following chemical equation:
$I_2+KOH \rightarrow KIO_3+KI+H_2O$

SOLUTION:
Using the principle of conservation of mass, we have:
${\color{red} a}I_2+{\color{red} b}KOH \rightarrow {\color{red} c}KIO_3+{\color{red} d}KI+{\color{red} e}H_2O$
$\begin{cases}K:b=c+d\\ I:2a=c+d\\ O:b=3c+e\\ H:b=2e\end{cases}$
There are 4 equations and 5 variables. We havo to find more relationships.

Is this a reduction-oxidation equation? Look at the oxidation number:
$\overset{0}{I_2}+KOH \rightarrow K\overset{+5}{I}+K\overset{-1}{I}+H_2O$
The iodine element $I_2$ changes its oxidation number from 0 to +5 (it is oxidating) and to -1 (it is reducing). At the same time, the iodine is the oxidant and the reductant. This is a disproportionation or dismutation equation.
But don't worry. We can use the same method to balance.

The two half-equations are:
$\begin{cases} \overset{0}{I_2} +2e^-\rightarrow 2\overset{-1}{I}\\\\ \overset{0}{I_2} \rightarrow 2\overset{+5}{I}+10e^-\end{cases}$
Now balance the electrons and sum the two half-equations:
$$\begin{cases} \overset{0}{I_2} +2e^-\rightarrow 2\overset{-1}{I} & \quad \mbox{(x5)}\\\\ \overset{0}{I_2} \rightarrow 2\overset{+5}{I}+10e^-\end{cases} \quad\quad+$$
$$\rule[0.3cm]{8cm}{0.01cm} $$
$$= \quad\quad 6\overset{0}{I_2}+10e^- \rightarrow 10\overset{-1}{I}+2\overset{+5}{I}+10e^-$$

So we have found that:
$\begin{cases} a=6\\ c=2\\ d=10\end{cases}$
From the relationships previously found, we can calculate:
$\begin{cases} b=12\\ e=6\end{cases}$

This dismutation reaction balanced is:
${\color{red} 6}I_2+{\color{red} 12}KOH \rightarrow {\color{red} 2}KIO_3+{\color{red} 10}KI+{\color{red} 6}H_2O$
All the stoichiometric coefficients we calculated are even. We can divide by two all of this:
${\color{red} 3}I_2+{\color{red} 6}KOH \rightarrow KIO_3+{\color{red} 5}KI+{\color{red} 3}H_2O$

---

PROBLEM 12:
Balance the following chemical equation:
$$Cu(CNS) \rightarrow Cu_2S+CS_2+(CN)_2+N_2$$
Continue reading »

How to balance redox equations #2

Balance the following chemical equation:
$$NaClO_4+NH_4Cl \rightarrow NaCl+N_2+HCl+H_2O$$

SOLUTION:
Balancing using the inspection technique, we have:
$${\color{red} a}NaClO_4+{\color{red} b}NH_4Cl \rightarrow {\color{red} c}NaCl+{\color{red} d}N_2+{\color{red} e}HCl+{\color{red} f}H_2O$$
$$\begin{cases}Na:a=c\\ Cl:a+b=c+e\\ N:b=2d\\ H:4b=e+2f\\ O:4a=f \end{cases}$$
We have 5 equation and 6 variable, so we cannot solve.

Try now to write down all oxidation numbers: is this a redox equation? Yes, in fact we have:
$$Na\overset{+7}{Cl}O_4+\overset{-3}{N}H_4Cl \rightarrow Na\overset{-1}{Cl}+\overset{0}{N_2}+H\overset{-1}{Cl}+H_2O$$
The chlorine changes its oxidation number from +7 to -1: it is reducing.
The nitrogen changes its oxidation number from -3 to 0: is is oxidating.

The two half-equations are
$$\begin{cases} \overset{+7}{Cl}+8e^- \rightarrow \overset{-1}{Cl}\\\\ 2\overset{-3}{N} \rightarrow \overset{0}{N_2}+6e^-\end{cases}$$
In the nitrogen equation we have $$6e^-$$ because there is the molecular compund $$N_2$$, and so we have to multiply by two.
Now balance the electrons and sum the two equation:
$$\begin{cases} \overset{+7}{Cl}+8e^- \rightarrow \overset{-1}{Cl} & \quad \mbox{(x3)}\\\\ 2\overset{-3}{N} \rightarrow \overset{0}{N_2}+6e^- & \quad \mbox{(x4)}\end{cases}\quad +$$
$$\rule[0.3cm]{8cm}{0.01cm} $$
$$= \quad3\overset{+7}{Cl}+8\overset{-3}{N} +24e^-\rightarrow 3\overset{-1}{Cl}+4\overset{0}{N_2}+24e^-$$

With the balancing of the oxidation numbers we have found the value of four stoichiometric coefficients:
$$\begin{cases}a=3\\ b=8\\ c=3 \\d=4\end{cases}$$
From the relationships found balancing for the principle of mass, we have:
$$\begin{cases}f=12\\ e=8\end{cases}$$

So the balanced equation is:
$${\color{red} 3}NaClO_4+{\color{red} 8}NH_4Cl \rightarrow {\color{red} 3}NaCl+{\color{red} 4}N_2+{\color{red} 8}HCl+{\color{red} 12}H_2O$$

---

PROBLEM 11:
Balance the following chemical equation:
$$(NH_4)_2Cr_2O_7 \rightarrow Cr_2O_3+N_2+H_2O$$
Continue reading »

How to balance redox equations

Balance the following redox equation using the inspection technique and the oxidation number method:
$$Ag_2O+H_2O_2 \rightarrow Ag+H_2O +O_2$$

SOLUTION:
First: try to balance this chemical equation using the inspection technique (using the principle of conservation of mass). We write down the stoichiometric coefficients:
$${\color{red} a}Ag_2O+{\color{red} b}H_2O_2 \rightarrow {\color{red} c}Ag+{\color{red} d}H_2O +{\color{red} e}O_2$$
For the principle of conservation of mass, we have:
$$\begin{cases}Ag:2a=c\\ H:2b=2d\\ O:a+2b=d+2e\end{cases}$$
We have five variables (from a to e) and only three equation. So we cannot balance correctly the equation only with the inspection technique.

Second: is this a redox equation? Write down all oxidation number.
$$\overset{+1}{Ag_2}\overset{-2}{O}+\overset{+1}{H_2}\overset{-1}{O_2} \rightarrow \overset{0}{Ag}+\overset{+1}{H_2}\overset{-2}{O}+\overset{0}{O_2}$$
As we can see, some atoms change their oxidation number: the Ag and the oxygen. So we can write the oxidating equation for the oxygen and the reducing equation for Ag:
$$\begin{cases}\overset{+1}{Ag_2}+xe^- \rightarrow 2\overset{0}{Ag}\\\\ \overset{-1}{O_2} \rightarrow \overset{0}{O_2}+ye^-\end{cases}$$
Now find the x and y numbers of electrons. There are two electrons in the oxidating and two in the reducting equation because there are two atoms of Ag and two atoms of O. So we have:
$$\begin{cases}\overset{+1}{Ag_2}+2e^- \rightarrow 2\overset{0}{Ag}\\\\ \overset{-1}{O_2} \rightarrow \overset{0}{O_2}+2e^-\end{cases}$$


Third: the electrons are balanced (there are 2 in either the equations). So we can now sum:
$$\begin{cases}\overset{+1}{Ag_2}+2e^- \rightarrow 2\overset{0}{Ag}\\\\ \overset{-1}{O_2} \rightarrow \overset{0}{O_2}+2e^-\end{cases}+$$
$$\rule[0.3cm]{5.5cm}{0.01cm} $$
$$=\quad \overset{+1}{Ag_2}+\overset{-1}{O_2} \rightarrow 2\overset{0}{Ag}+\overset{0}{O_2}$$

So we have found 4 variable values:
$$\begin{cases} a=1\\b=1\\c=2\\e=1\end{cases}$$
From the equation of the hydrogen balancing, we found that $d=1$. We can verify this stoichiometric coefficients substituting the values in the oxygen equation of mass balancing.

So the balanced equation is:
$${\color{red} 1}Ag_2O+{\color{red} 1}H_2O_2 \rightarrow {\color{red} 2}Ag+{\color{red} 1}H_2O +{\color{red} 1}O_2$$
Or:
$$Ag_2O+H_2O_2 \rightarrow {\color{red} 2}Ag+H_2O +O_2$$

---

PROBLEM 10:
Balance the following reaction: manganese dioxide + hydrochloric acid produces manganese(II) ions + chlorine gas
Continue reading »

How to balance ionic equations

Balance this ionic equation: $SiF_6^{2-}+OH^-\to SiO_3^{2-}+F^-+H_2O$.

SOLUTION:
Write down the stoichiometric coefficients in letters:
${\color{red} a}SiF_6^{2-}+{\color{red} b}OH^-\to {\color{red} c}SiO_3^{2-}+{\color{red} d}F^-+{\color{red} e}H_2O$

Now balance considering the principle of conservation of mass, and the principle of conservation of charge (the total charge in reactants will be equal to the total charge in products):
$\begin{cases}Si:a=c\\ F:6a=d\\ O:b=3c+e\\ H:b=2e\\ Charge:2a+b=2c+d\end{cases}$

Try now to express every variable in terms of $a$. From the silicon equation we obtain $c=a$; from the fluorine equation we obtain $d=6a$. Now look at the oxygen and hydrogen equations:
$\begin{cases}O:b=3c+e\\ H:b=2e\\ c=a\end{cases}\Rightarrow 3c+e=2e\Rightarrow e=3c\Rightarrow e=3a$
And finally $b=2e\Rightarrow b=6a$. To verify, we can try to substitute the value of $b$, $c$, $d$ and $e$ (in terms of $a$) in the charge equation (if correct, we will obtain $a=a$).
So we have:
$\begin{cases}a=a\\ b=6a\\ c=a\\ d=6a\\ e=3a\end{cases}$

Now assign $a=1$. The balanced equation is:
$SiF_6^{2-}+{\color{red} 6}OH^-\to SiO_3^{2-}+{\color{red} 6}F^-+{\color{red} 3}H_2O$

---

PROBLEM 9:
When dissolved beryllium chloride reacts with dissolved silver nitrate in water, acqueous beryllium nitrate and silver chloride powder are made. Write the balanced equation and put it in the ionic form.
Continue reading »

How to balance acid-base equations #2

Balance this acid-base equation: $H_3AsO_4+P_2O_5\to H_3PO_4+As_2O_5$.

SOLUTION:
Start writing the chemical equation with letters indicating the stoichiometric coefficients:
${\color{red}a }H_3AsO_4+{\color{red}b }P_2O_5\to {\color{red}c }H_3PO_4+{\color{red}d }As_2O_5$
Now balance every atom species for the principle of conservation of mass:
$\begin{cases}As:a=2d\\ P:2b=c\\ O:4a+5b=4c+5d\\ H:3a=3c\end{cases}$

Try now to express every variable in terms of $a$. From the arsenic equation, we know that $d=\frac{1}{2}a$; from the hydrogen equation we know that $c=a$.
From the phosphorus equation we have: $2b=c=a\Rightarrow 2b=a\Rightarrow b=\frac{1}{2}a$. To verify, we can try to substitute the value of $b$, $c$ and $d$ (in terms of $a$) in the oxygen equation (if correct, we will obtain $a=a$).
$\begin{cases}a=a\\ b=\frac{1}{2}a\\ c=a\\ d=\frac{1}{2}a\end{cases}$

Now give to the stoichiometric coefficient $a$ the value $2$. So we have:
$\begin{cases}a=2\\ b=1\\ c=2\\ d=1\end{cases}$

The balanced chemical equation is:
${\color{red}2 }H_3AsO_4+{\color{red}1 }P_2O_5\to {\color{red}2 }H_3PO_4+{\color{red}1 }As_2O_5$
That is:
${\color{red}2 }H_3AsO_4+P_2O_5\to {\color{red}2 }H_3PO_4+As_2O_5$

---

PROBLEM 8:
Try to balance these two chemical equation:
$H_2SO_4+B(OH)_3 \rightarrow B_2(SO_4)_3+H_2O$
$ZnSO_4+Li_2CO_3 \rightarrow ZnCO_3+Li_2SO_4$
Continue reading »

How to balance acid-base equations

Balance this acid-base equation: $Ca(H_2PO_4)_2\to Ca(PO_3)_2+H_2O$.

SOLUTION:
Remember that the number of every single atom in reactants will be the same to the number of that atom in products (law of conservation of mass). Start writing the chemical equation with letters indicating the stoichiometric coefficients:
${\color{red}a }Ca(H_2PO_4)_2\to {\color{red} b}Ca(PO_3)_2+{\color{red} c}H_2O$
For the principle of conservation of mass, the number of $Ca$ in reactants would be equal to the number of $Ca$ in products: so $a=b$. Similarly we obtain other equations for the other atoms:
$\begin{cases}Ca:a=b\\ P:2a=2b\\ O:8a=6b+c\\ H:4a=2c\end{cases}$

Try now to express every variable in terms of $a$. From the calcium equation we know that $b=a$; from the hydrogen equation we know that $c=2a$. To verify this number, we can try to substitute the value of $b$ and $c$ (in terms of $a$) in the oxygen equation (if correct, we will obtain $a=a$).
$\begin{cases}a=a\\ b=a\\ c=2a\end{cases}$

Now give to the stoichiometric coefficient $a$ the value $1$. So we have:
$\begin{cases}a=1\\ b=1\\ c=2\end{cases}$

The balanced chemical equation is:
${\color{red}1 }Ca(H_2PO_4)_2\to {\color{red} 1}Ca(PO_3)_2+{\color{red} 2}H_2O$
That is:
$Ca(H_2PO_4)_2\to Ca(PO_3)_2+{\color{red} 2}H_2O$

---

PROBLEM 7:
Try to balance these two chemical equation:
$NaNO_3+PbO \rightarrow Pb(NO_3)_2+Na_2O$
$AgI+Fe_2(CO_3)_3 \rightarrow FeI_3+Ag_2CO_3$
Continue reading »

Find the empirical formula from a reaction

Silicon reacts with $H_2$, to form a compound with formula $Si_xH_y$. From the combustion of 6.22gr of $Si_xH_y$ we obtain 11.64gr of $SiO_2$ and 6.98gr of $H_2O$. What is the empirical formula of the compound?

SOLUTION:
First we can calculate the number of mole of $SiO_2$ and $H_2O$:
$$MW_{SiO_2}=60gr/mol$$
$$MW_{H_2O}=18gr/mol$$
$$n_{SiO_2}=\frac{11.64gr}{60gr/mol}=0.194mol$$
$$n_{H_2O}=\frac{6.98gr}{18gr/mol}=0.388mol$$

Every mole of $SiO_2$ comes from one mole of $Si$; every mole of $H_2O$ comes from two mole of $H$:
$$n_{Si}=n_{SiO_2}=0.194mol$$
$$n_H=2\cdot n_{H_2O}=0.776mol$$

Now we can calculate the mass (m) of Si and H in that number of moles:
$$m_{Si}=n_{Si}\cdot Si=0.194mol \cdot 28.09gr/mol=5.45gr$$
$$m_H=n_H \cdot H=0.776mol \cdot 1.01gr/mol=0.784gr$$

The percentage composition of $Si_xH_y$ is:
$$\%Si=\frac{5.45gr}{6.22gr}\cdot 100=89.6\%$$
$$\%H=\frac{0.784gr}{6.22gr}\cdot 100=12.6\%$$

In 100gr of $Si_xH_y$ there are 87.6gr of Si and 12.6gr of H; the relative numbers of mole are:
$$n_{Si}=\frac{87.6}{28.09}=3.12$$
$$n_H=\frac{12.6}{1.01}=12.47$$

Now we can calculate the empirical formula of $Si_xH_y$:
$$x:y=3.12:12.47=\frac{3.12}{3.12}:\frac{12.47}{3.12}=1:3.99$$

The empirical formula is $$SiH_4$$.

---

PROBLEM 6:
0.520 grams of $NiSO_4 \cdot xH_2O$ are heated to drive off the water. The reaction gave a residue of 0.306gr. Find the formula of the hydrate.
Continue reading »

How to find the empirical formula from mass percentage

Elemental analysis of a compound showed that it consisted of: $52.17\%C$, $13.05\%H$, $34.78\%O$. Find the empirical formula of the compound.

SOLUTION:
We can assume that 100gr of the substance contain 52.17gr of carbon, 13.05gr of hydrogen, and 34.78gr of oxygen.
Remembering the atomic weight, we can calculate the number of mole (n):

$$n_C=\frac{52.17gr}{12gr/mol}=4.35mol$$

$$n_H=\frac{13.05gr}{1.01gr/mol}=12.9mol$$

$$n_O=\frac{34.78gr}{16gr/mol}=2.17mol$$

The empirical formula is $C_xH_yO_z$; so we have:
$x:y:z=4.35:12.9:2.17$

We divide now by the smallest number (oxygen), to obtain:
$$x:y:z=\frac{4.35}{2.17}:\frac{12.9}{2.17}:\frac{2.17}{2.17}=2:6:1$$

So, the empirical formula is $C_2H_6O$.

---

PROBLEM 5:
Determine the empirical formula of a compound that showed the following analytical results: $26.6\%K$, $35.4\%Cr$, $38.0\%O$.
Continue reading »

Calculation of formula from chemical analysis

The percentage composition of an unknown compound is: $27.05\%Na$, $16.48\%N$, $56.47\%O$. Find the empirical formula of the compound.

SOLUTION:
First, remember the atomic weight of Na, N, O:
$$\begin{cases}Na=23\\ N=14\\ O=16\end{cases}$$
We can say that if we have 100gr of the compound, 27.05gr is Na, 16.48gr is N and 56.47gr is O. Now we can found the number of mole (n) of sodium, nitrogen and oxygen in 100gr of the compound:

$$n_{Na}=\frac{27.05gr}{23gr/mol}=1.18mol$$

$$n_N=\frac{16.48gr}{14gr/mol}=1.18mol$$

$$n_O=\frac{56.47gr}{16gr/mol}=3.54mol$$

The empirical formula is $Na_xN_yO_z$;
$x:y:z=1.18:1.18:3.54$
Dividing by the smallest number, we find:
$$x:y:z=\frac{1.18}{1.18}:\frac{1.18}{1.18}:\frac{3.54}{1.18}=1:1:3$$

So, the empirical formula is $NaNO_3$.

---

PROBLEM 4:
From the following analytical results (percentage by weight), determine the empirical formula fro the compound analyzed: $21.8\%Mg$, $27.9\%P$, $50.3\%O$.
Continue reading »

Calculate molecular weight composition

We know that 450mg of sodium are contained in a certain quantity of $Na_2S_2O_3$. How many grams of sulfur are there in the same quantity?

SOLUTION:
The molecular weight of $Na_2S_2O_3$ is:
$2\cdot23+2\cdot32.07+3\cdot16=158.14gr/mol$
The sodium percentage is:
$$\%Na=\frac{2\cdot23gr/mol}{158.14gr/mol}\cdot100=29.1\%$$

So, in 100gr of $Na_2S_2O_3$ there are 29.1gr of sodium; in $X$ grams of $Na_2S_2O_3$ there are 0.45gr of sodium:
$$100gr:29.1gr=X:0.45gr\Rightarrow X=\frac{0.45\cdot100}{29.1}=1.55gr$$

The sulfur percentage is:
$$\%S=\frac{2\cdot32.07gr/mol}{158.14}\cdot100=40.55\%$$

So, in 100gr of $Na_2S_2O_3$ there are 40.55gr of sulfur; in 1.55gr of $Na_2S_2O_3$ there are $X$ grams of sulfur:
$$100gr:40.55gr=1.55gr:X\Rightarrow X=\frac{40.55\cdot1.55}{100}=0.628gr=628mg$$

---

PROBLEM 3:
3.5 grams of aluminium are contained in a certain quantity of aluminium sulfate. How many grams of oxygen are contained in the same quantity of the compound?
Continue reading »

Calculate molecular weight and the percent composition

Calculate the mass (in gr) of $Pb_3(PO_4)_3$ that contains $0.3gr$ of $Pb$. Then calculate the mass of $P$ contained in the same quantity of $Pb_3(PO_4)_3$. What is the percentage of oxygen?

SOLUTION:
Remember the atomic weight of the single atoms:
$\left\{\begin{matrix} Pb=207.2\\ P=31\\ O=16\\ \end{matrix}\right.$

The molecular weight of $Pb_3(PO_4)_3$ is:
$MW=207.2\cdot 3+\left ( 31+16\cdot 4 \right )\cdot 2=811.6gr/mol$
So one mole of $Pb_3(PO_4)_3$ has a mass of $811.6gr$.

The percentage of $Pb$ is:
$\%Pb=\frac{3\cdot 207.2gr/mol}{\displaystyle 811.6gr/mol}\cdot 100=76.59\%$
So, in $100gr$ of $Pb_3(PO_4)_3$ there are $76.59gr$ of $Pb$; $X$ grams of $Pb_3(PO_4)_3$ contains $0.3gr$ of $Pb$. Then:
$100gr:76.59gr=X:0.3gr\Rightarrow X=\frac{100gr\cdot 0.3gr}{76.59gr}=0.392gr=392mg$

The percentage of $P$ is:
$\%Pb=\frac{2\cdot 31gr/mol}{811.6gr/mol}\cdot 100=7.64\%$
So, in $100gr$ of $Pb_3(PO_4)_3$ there are $7.64gr$ of $P$; $0.392gr$ of $Pb_3(PO_4)_3$ contains $X$ grams of $P$. Then:
$100gr:7.64gr=0.392gr:X\Rightarrow X=\frac{7.64gr\cdot 0.392gr}{100gr}=0.0299gr=29.9mg$

The percentage of oxygen is:
$\%O=\frac{4\cdot 2\cdot 16gr/mol}{811.6gr/mol}\cdot 100=15.7\%$

---

PROBLEM 2:
Calculate the mass of barium sulfate that contains 1.2gr of sulfur. Then compute the percentage of oxygen.
Continue reading »

How to calculate the percentage composition

Calculate the composition in percentage of the single atoms and $H_2O$ in the chemical compound $Na_2B_4O_7\cdot 10H_2O.$

SOLUTION:
First we consider the atomic weight of the species:
$\begin{cases} Na=23\\ B=10.8\\ O=16\\ H_2O=18\\ \end{cases}$

The molecular weight of $Na_2B_4O_7\cdot 10H_2O$ is:

$MW=2\cdot 23+4\cdot 10.8+7\cdot 16+10\cdot 18=381.2gr/mol$
So one mole of $Na_2B_4O_7\cdot 10H_2O$ has a mass of 381.2gr.

The percentage of the atoms in the compound is, in general:
$\%X=\frac{nX\cdot AtomicWeight}{AtomicMass}$

We can now calculate:
$\%Na=\frac{2\cdot 23gr/mol}{381.2gr/mol}\cdot 100=12.06\%$

$\%B=\frac{4\cdot 10.8gr/mol}{381.2gr/mol}\cdot 100=11.33\%$

$\%O=\frac{7\cdot 16gr/mol}{381.2gr/mol}\cdot 100=29.38\%$

$\%H_2O=\frac{10\cdot 18gr/mol}{381.2gr/mol}\cdot 100=47.22\%$

---

PROBLEM 1:
Compute the percentage of K, Fe, C, N and $H_2O$ in $K_4Fe(CN)_6 \cdot 5H_2O$ crystals. Continue reading »

Exercise Index

Chapter 1: chemical compounds and formulas

1. Atomic and molecular weights: Problem 1
2. Chemical composition: weight and volume percent: Problem 1; Problem 2
3. Chemical analysis and compound's formula: Problem 1; Problem 2; Problem 3
   

Chapter 2: balancing equations

1. How to balance acid-base equations: Problem 1; Problem 2
   
2. How to balance redox equations: Problem 1; Problem 2; Problem 3; Problem 4
   
3. How to balance ionic equations: Problem 1
4. How to balance redox equation in ionic form: Problem 1; Problem 2; Problem 3; Problem 4
   
5. How to balance electrodic equations:
   

Chapter 3: Stoichiometry calculations

1. Gas formation at standard conditions from reactions: Problem 1; Problem 2; Problem 3
2. Chemical equations and stoichiometry calculations: Problem 1; Problem 2; Problem 3
   

Continue reading »