Chapter 9, Problem 33E

Interpretation Introduction

Interpretation:

The milliliters of carbon dioxide gas at STP that are produced from the decomposition of $\text{5}\text{.00}\text{g}$ of iron $\left(\text{III}\right)$ carbonate are to be calculated.

Concept introduction:

Chemical reactions are represented by chemical equations. In a chemical equation, the reactants are represented on the left of the arrow while the products are represented on the right of the arrow. Stoichiometric coefficient is the number preceding each symbol in a reaction which determines the moles of the reactants and products in the reaction. The ratio of moles is termed as mole ratio. In stoichiometry problems, a known amount of product or reactant is given and the other product or reactant has to be calculated.

Answer to Problem 33E

The milliliters of carbon dioxide gas at STP that are produced from the decomposition of $\text{5}\text{.00}\text{g}$ of iron $\left(\text{III}\right)$ carbonate are $1151.75\text{mL}$.

Explanation of Solution

The reaction is given below.

${\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3\left(s\right)\stackrel{\Delta }{\to }{\text{Fe}}_2{\text{O}}_{\text{3}}\left(s\right)+3{\text{CO}}_{\text{2}}\left(g\right)$

In the reaction, $1$ mole of ${\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3$ produces $3$ moles of ${\text{CO}}_{\text{2}}$.

Therefore, the mole ratio is given below.

$\frac{1\text{mol}{\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3}{3\text{mol}{\text{CO}}_{\text{2}}}\text{and}\frac{3\text{mol}{\text{CO}}_{\text{2}}}{1\text{mol}{\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3}$

The mole ratio to obtain moles of ${\text{CO}}_{\text{2}}$ from the moles of ${\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3$ is given below.

$\frac{3\text{mol}{\text{CO}}_{\text{2}}}{1\text{mol}{\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3}$

The molar mass of iron is $55.85\text{g}{\text{mol}}^{-1}$.

The molar mass of oxygen is $16.00\text{g}{\text{mol}}^{-1}$.

The molar mass of carbon is $12.01\text{g}{\text{mol}}^{-1}$

Therefore, the molar mass of ${\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \left(2\times 55.85\text{g}{\text{mol}}^{-1}\right)+3\times \left(12.01\text{g}{\text{mol}}^{-1}+\left(3\times 16.00\text{g}{\text{mol}}^{-1}\right)\right)\\ & =& 111.7\text{g}{\text{mol}}^{-1}+3\times \left(12.01\text{g}{\text{mol}}^{-1}+48.00\text{g}{\text{mol}}^{-1}\right)\\ & =& 111.7\text{g}{\text{mol}}^{-1}+180.03\text{g}{\text{mol}}^{-1}\\ & =& 291.73\text{g}{\text{mol}}^{-1}\end{array}$

The unit factor to calculate moles of ${\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3$ from grams of ${\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3$ is shown below.

$\frac{\text{1}\text{mol}{\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3}{291.73\text{g}{\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3}$

The volume of $\text{1}\text{mol}$ of gas at STP is $\text{22400}\text{mL}$.

The formula that is used to calculate the volume of ${\text{CO}}_{\text{2}}$ from the mass of ${\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3$ is given below.

$\text{Volume}\text{of}{\text{CO}}_{\text{2}}=\left(\begin{array}{l}\text{Given}\text{mass}\text{of}{\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3\\ \times \text{unit factor}\text{to}\text{obtain}\text{moles}\text{of}{\text{CO}}_{\text{2}}\\ \times \text{unit factor}\text{to}\text{obtain}\text{moles}\text{of}{\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3\\ \times \frac{\text{22400}\text{mL}}{1\text{mol}{\text{CO}}_{\text{2}}}\end{array}\right)$…(1)

The mass of ${\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3$ is $\text{5}\text{.00}\text{g}$.

Substitute the value of mass of ${\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3$, mole ratio and unit factors in equation (1).

$\begin{array}{rll}\text{Volume}\text{of}{\text{CO}}_{\text{2}}& =& \left(\begin{array}{l}\text{5}\text{.00}\text{g}\text{of}{\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3\times \frac{3\text{mol}{\text{CO}}_{\text{2}}}{1\text{mol}{\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3}\times \frac{\text{1}\text{mol}{\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3}{291.73\text{g}{\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3}\\ \times \frac{\text{22400}\text{mL}}{1\text{mol}{\text{CO}}_{\text{2}}}\end{array}\right)\\ & =& \text{5}\text{.00}\text{g}\text{of}{\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3\times 3\times \frac{1}{291.73\text{g}{\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3}\times \text{22400}\text{mL}\\ & =& 1151.74\text{mL}\end{array}$

Therefore, the milliliters of carbon dioxide gas at STP that are produced from the decomposition of $\text{5}\text{.00}\text{g}$ of iron $\left(\text{III}\right)$ carbonate are $1151.75\text{mL}$.

Conclusion

The milliliters of carbon dioxide gas at STP that are produced from the decomposition of $\text{5}\text{.00}\text{g}$ of iron $\left(\text{III}\right)$ carbonate are $1151.75\text{mL}$.