Question

Write a balanced equation for the combustion of benzoic acid,${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{CO}}_{\mathbf{2}}\mathbf{H}$, to give${\mathbf{CO}}_{\mathbf{2}}$ and ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$. How many milligrams of and of will be produced by the combustion of 4.635mgof benzoic acid?

Short answer

The mass of ${\mathrm{CO}}_2$and ${\mathrm{H}}_2\mathrm{O}$produced during the combustion of 4.635mg of benzoic acid is 0.01169 and $2.051\times {10}^{-3}$grespectively.

Step-by-step solution

Writing a balanced equation.

A balanced equation is a chemical reaction equation in which the total charge and the number of atoms for each element in the reaction are the same for both the reactants and the products. In other words, the mass and charge on both sides of the reaction are balanced.

Calculating the moles of benzoic acid.

Consider the following balanced equation of benzoic acid combustion:

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CO}}_2\mathrm{H}+\frac{15}{2}{\mathrm{O}}_2\to 7{\mathrm{CO}}_2+3{\mathrm{H}}_2\mathrm{O}$

To write a balanced equation for the combustion of benzoic acid to give and and calculate the mass of products given by the combustion of of benzoic acid.

Calculating the moles of benzoic acid by the following:

$$\begin{gathered} \mathrm{n}\left(\mathrm{benzoic}\mathrm{acid}\right)=\frac{\mathrm{m}}{\mathrm{M}} \\ \mathrm{n}\left(\mathrm{benzoic}\mathrm{acid}\right)=\frac{4.635\times {10}^{-3}\mathrm{g}}{122.123\mathrm{g}/\mathrm{mol}} \\ \mathrm{n}\left(\mathrm{benzoic}\mathrm{acid}\right)=3.795\times {10}^{-5}\mathrm{mol} \end{gathered}$$

Calculating the mass of products.

Next calculate the mass of products:

Carbondioxide:

$$\begin{gathered} \mathrm{n}\left({\mathrm{CO}}_2\right)=7\mathrm{n}\left(\mathrm{benzoic}\mathrm{acid}\right) \\ \mathrm{n}\left({\mathrm{CO}}_2\right)=7\times \left(3.795\times {10}^{-5}\mathrm{mol}\right) \\ \mathrm{n}\left({\mathrm{CO}}_2\right)=2.6565\times {10}^{-4}\mathrm{mol} \end{gathered}$$

1 mole of benzoic acid gives 7 moles of carbon dioxide

$$\begin{gathered} \mathrm{m}\left({\mathrm{CO}}_2\right)=\mathrm{n}\times \mathrm{M} \\ \mathrm{m}\left({\mathrm{CO}}_2\right)=\left(2.6565\times {10}^{-4}\mathrm{mol}\right)\times 44.010\mathrm{g}/\mathrm{mol} \\ \mathrm{n}\left({\mathrm{CO}}_2\right)=0.01169\mathrm{g} \end{gathered}$$

Water:

role="math" localid="1663662527422" $$\begin{gathered} \mathrm{n}\left({\mathrm{H}}_2\mathrm{O}\right)=3\mathrm{n}\left(\mathrm{benzoic}\mathrm{acid}\right) \\ \mathrm{n}\left({\mathrm{H}}_2\mathrm{O}\right)=3\times \left(3.795\times {10}^{-5}\mathrm{mol}\right) \\ \mathrm{n}\left({\mathrm{H}}_2\mathrm{O}\right)=1.1385\times {10}^{-4}\mathrm{mol} \end{gathered}$$

1 mole of benzoic acid gives 3 moles of water

role="math" localid="1663662547240" $$\begin{gathered} \mathrm{m}\left({\mathrm{H}}_2\mathrm{O}\right)=\mathrm{n}\times \mathrm{M} \\ \mathrm{m}\left({\mathrm{H}}_2\mathrm{O}\right)=\left(1.1385\times {10}^{-4}\mathrm{mol}\right)\times 18.015\mathrm{g}/\mathrm{mol} \\ \mathrm{n}\left({\mathrm{H}}_2\mathrm{O}\right)=2.051\times {10}^{-3}\mathrm{g} \end{gathered}$$

Therefore the mass of ${\mathrm{CO}}_2$and ${\mathrm{H}}_2\mathrm{O}$ produced during the combustion of 4.635mg of benzoic acid is 0.01169g and$2.051\times {10}^{-3}$ respectively.