Chapter 9, Problem 34E

Interpretation Introduction

Interpretation:

The grams of iron $\left(\text{III}\right)$ carbonate that decompose to give $\text{50}\text{.0}\text{mL}$ of carbon dioxide gas at STP 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 34E

The grams of iron $\left(\text{III}\right)$ carbonate that decompose to give $\text{50}\text{.0}\text{mL}$ of carbon dioxide gas at STP are $0.216\text{g}$.

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 ${\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3$ from moles of ${\text{CO}}_{\text{2}}$ is given below.

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

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

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

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

At STP $22400\text{mL}$ is occupied by $1\text{mol}$ of ${\text{CO}}_{\text{2}}$.

Therefore, the number of moles of ${\text{CO}}_{\text{2}}$ occupying $\text{50}\text{.0}\text{mL}$ is calculated below.

$\begin{array}{rll}\text{Moles}\text{of}{\text{CO}}_{\text{2}}& =& \frac{50.0}{22400}\text{mol}\\ & =& 2.23\times {10}^{-3}\text{mol}\end{array}$

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

$\text{Mass of} {\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3=\left(\begin{array}{l}\text{Moles}\text{of}{\text{CO}}_{\text{2}}\\ \times \text{unit}\text{factor}\text{to}\text{obtain}\text{moles}\text{of}{\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3\\ \times \text{unit}\text{factor}\text{to}\text{obtain}\text{grams}\text{of}{\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3\end{array}\right)$…(1)

Substitute the value of moles of ${\text{CO}}_{\text{2}}$, the mole ratio and unit factors in equation (1).

$\begin{array}{rll}\text{Mass}\text{of}{\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3& =& \left(2.23\times {10}^{-3}\text{mol}{\text{CO}}_{\text{2}}\times \frac{1\text{mol}{\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3}{3\text{mol}{\text{CO}}_{\text{2}}}\times \frac{291.73\text{g}{\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3}{\text{1}\text{mol}{\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3}\right)\\ & =& 2.23\times {10}^{-3}\text{mol}{\text{CO}}_{\text{2}}\times \frac{1}{3\text{mol}{\text{CO}}_{\text{2}}}\times 291.73\text{g}{\text{Fe}}_2{\left({\text{CO}}_{\text{3}}\right)}_3\\ & =& 0.216\text{g}\end{array}$

Therefore, the grams of iron $\left(\text{III}\right)$ carbonate that decompose to give $\text{50}\text{.0}\text{mL}$ of carbon dioxide gas at STP are $0.216\text{g}$.

Conclusion

The grams of iron $\left(\text{III}\right)$ carbonate that decompose to give $\text{50}\text{.0}\text{mL}$ of carbon dioxide gas at STP are $0.216\text{g}$.