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:
${\color{red}2 }H_3AsO_4+P_2O_5\to {\color{red}2 }H_3PO_4+As_2O_5$

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PROBLEM 8:
Try to balance these two chemical equation:
$H_2SO_4+B(OH)_3 \rightarrow B_2(SO_4)_3+H_2O$
$ZnSO_4+Li_2CO_3 \rightarrow ZnCO_3+Li_2SO_4$
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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:
$Ca(H_2PO_4)_2\to Ca(PO_3)_2+{\color{red} 2}H_2O$

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