## Monday, August 3, 2009

### Stoichiometry calculations #1

How many grams of water and carbon dioxide are formed after the combustion of 1.275 grams of sucrose?

SOLUTION:
Recall that the chemical formula of sucrose is: $C_{12}H_{22}O_{11}$, and its molecular weight is: $MW=342.34gr/mol$.

The equation of combustion of sucrose is as follows:
$$C_{12}H_{22}O_{11}+O_2 \rightarrow CO_2+H_2O$$

This reaction must be balanced first.
Being a redox reaction, we can write the two separate reactions of oxidation and reduction:

$$\overset{0}{C_{12}}\overset{+1}{H_{22}}\overset{-2}{O_{11}}+\overset{0}{O_2} \rightarrow \overset{+4}{C}\overset{-2}{O_2}+\overset{+1}{H_2}\overset{-2}{O}$$

$$\begin{cases} \overset{0}{O_2}+4e^- \rightarrow \overset{-2}{O_2} & \quad \mbox{(x12)}\\\\ \overset{0}{C_{12}} \rightarrow 12\overset{+4}{C}+12 \cdot 4e^- \end{cases}\quad +$$
$$\rule[0.3cm]{8cm}{0.01cm}$$
$$= \quad \overset{0}{C_{12}}+12\overset{0}{O_2} \rightarrow 12\overset{+4}{C}\overset{-2}{O_2}$$

The balanced equation is:

$$C_{12}H_{22}O_{11}+12O_2 \rightarrow 12CO_2+11H_2O$$

The number of moles of sucrose contained in 1.275 grams is:

$$n_{C_{12}H_{22}O_{11}}=\frac{1.275gr}{342.34gr/mol}=0.0037mol$$

From the stoichiometry of the reaction, we know that for every mole of sucrose that is burned, you obtain 12 moles of carbon dioxide and 11 moles of water. Then we have:

$$n_{CO_2}=12n_{C_{12}H_{22}O_{11}}=0.0447mol$$

$$n_{H_2O}=11n_{C_{12}H_{22}O_{11}}=0.0410mol$$

Knowing the molecular weight of carbon dioxide (44gr/mol) and water (18gr/mol), we can calculate the weight (in grams) that is formed for each compound:

$$m_{CO_2}=44gr/mol \cdot 0.0447mol=1.9668gr$$

$$m_{H_2O}=18gr/mol \cdot 0.0410mol=0.7380gr$$

PROBLEM 21:
What mass of calcium carbonate would be formed if 248.6g of carbon dioxide were exhaled into limewater, $Ca(OH)_2$? How many grams of calcium would be needed to form that amount of calcium carbonate? Assume 100% yield in each reaction.

## Sunday, August 2, 2009

### Gas formation from reaction #3

A mixture of iron (II) sulfide contains 93% of iron sulfide and 3% of metallic iron. Compute the volume (at standard condition) of gas that is formed when 10 grams of the mixture react with acid. The two reactions that occur are:
$$FeS+H^+ \rightarrow Fe^{2+}+H_2S$$
$$Fe+H^+ \rightarrow Fe^{2+}+H_2$$

SOLUTION:
The first thing to do is balance the two reactions:
$$FeS+{\color{red} 2}H^+ \rightarrow Fe^{2+}+H_2S$$
$$Fe+{\color{red} 2}H^+ \rightarrow Fe^{2+}+H_2$$

Since the 10 grams of the mixture are 97% of FeS and 3% for Fe, we can calculate the amount in grams of each component, and, knowing the molecular weight, calculate the number of moles, as follows:

$$97\%(10gr)=9.7gr(FeS)$$
$$3\%(10gr)=0.3gr(Fe)$$

Number of moles:
$$n_{FeS}=\frac{m_{FeS}}{MW_{FeS}}=\frac{9.7gr}{(55.85+32.07)gr/mol}=0.11mol$$

$$n_{Fe}=\frac{m_{Fe}}{MW_{Fe}}=\frac{0.3gr}{55.85gr/mol}=0.005mol$$

Since both the hydrogen sulfide and pure hydrogen are gas, we must calculate the number of moles of both species. Knowing the stoichiometry of the two reactions, we deduce that:

1) for each mole of Fes that reacts, it forms one mole of $H_2S$:

$$n_{H_2S}=n_{FeS}=0.11mol$$

Since a mole of gas is 22.4 liters, we have that the volume of H_2S that is formed is equal to:
$$V_{H_2S}=n_{H_2S}=0.11mol \cdot 22.4L/mol=2.464L$$

2) for each mole of iron metal that reacts, it forms a mole of hydrogen gas:

$$n_{H_2}=n_{Fe}=0.005mol$$
$$V_{H_2}=0.005mol \cdot 22.4L/mol=0.112L$$

The total volume of gas that is formed is then:

$$V_{tot}=V_{H_2S}+V_{H_2}=2.576L$$

PROBLEM 20:
how many liters of chlorine are required to transform 50.5 grams of $PCl_3$ in $PCl_5$?

## Saturday, August 1, 2009

### Gas formation from reaction #2

Compute the oxygen volume (at standard conditions: 0°C, 1atm) that you need to burn 76.14gr of carbon disulfide in carbon dioxide and sulphur dioxide. Then compute the volume of carbon dioxide and sulphur dioxide and their percentage composition.

SOLUTION:
The balanced reaction that take place is:
$$CS_2(l)+{\color{red} 3}O_2(g) \rightarrow CO_2(g)+{\color{red} 2}SO_2(g)$$

For every mole of $CS_2$ that burns, we need of 3 moles of oxygen. Then:
$$n_{CS_2}=\frac{m_{CS_2}}{MW_{CS_2}}=\frac{76.14gr}{76.14gr/mol}=1mol$$

$$n_{O_2}=3n_{CS_2}$$
$$V_{O_2}=n_{O_2}\cdot22.4=67.2L$$

The volume of carbon dioxide correspond to one mole: 22.4L
The volume of sulpgur dioxide correspond to two moles: 44.8L

The percent composition of $CO_2$ and $SO_2$ is:

$$\%CO_2=\frac{n_{CO_2}}{n_{CO_2}+n_{SO_2}}\cdot100=\frac{1}{1+2}\cdot100=33\%$$

$$\%SO_2=\frac{n_{SO_2}}{n_{CO_2}+n_{SO_2}}\cdot100=\frac{2}{1+2}\cdot100=67\%$$

PROBLEM 19:
1,00 gram of a mixture of $CaCO_3$ and $MgCO_3$ gives 240mL of $CO_2$ after heating. Compute the percent composition of the mixture (remember that magnesium carbonate and calcium carbonate decompose on heating in the corresponding oxide and carbon dioxide).

## Friday, July 31, 2009

### 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 $O_2$.

## Thursday, July 30, 2009

### 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:
$$[Ni(NH_3)_4]Cl_2+(C_6H_5)NC+H_2O \rightarrow$$
$$\quad Ni[(C_6H_5)NC]_4+NH_3+NH_4Cl+(C_6H_5)NCO$$