Chapter 15, Problem 26E

Interpretation Introduction

(a)

Interpretation:

The stoichiometry concept map is to be drawn and the mass of ${\text{KHCO}}_{\text{3}}$ decomposed is to be stated.

Concept introduction:

Stoichiometry measures the amount of reactant and product consumed or formed in a chemical reaction. It gives the relationship between the reactants and products. Stoichiometry concept maps are used to determine the relation of reactant and product in terms of stoichiometry.

Answer to Problem 26E

The stoichiometry concept map is shown below.

The mass of ${\text{KHCO}}_{\text{3}}$ decomposed is $2.282\text{g}$.

Explanation of Solution

It is given that $\text{255}\text{mL}$ of carbon dioxide is produced at $\text{STP}$.

The given chemical reaction is stated below.

${\text{KHCO}}_3(s)\to {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}(aq)+{\text{H}}_{\text{2}}\text{O}(l)+{\text{CO}}_{\text{2}}\left(g\right)$…(1)

The stoichiometry concept map is shown below.

Figure 1

The chemical reaction (1) should be balanced first to calculate the mass of ${\text{KHCO}}_{\text{3}}$ decomposed.

The number of potassium atom and the hydrogen atom are not equal on both sides. A coefficient of $2$ is added in front of ${\text{KHCO}}_3$ to balance the chemical reaction.

The balanced equation is stated below.

${\text{2KHCO}}_3(s)\to {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}(aq)+{\text{H}}_{\text{2}}\text{O}(l)+{\text{CO}}_{\text{2}}\left(g\right)$…(2)

The equation (2) shows that $2$ mole of ${\text{KHCO}}_{\text{3}}$ is consumed to form $1$ mole of carbon dioxide gas.

At $\text{STP}$ the relation between ${\text{KHCO}}_{\text{3}}$ and carbon dioxide gas is shown below.

$\text{1}\text{mole}\text{of}{\text{KHCO}}_{\text{3}}=\text{22}\text{.4}\text{L}{\text{CO}}_2$

Therefore, the number of moles of ${\text{KHCO}}_{\text{3}}$ is calculated as shown below.

$\begin{array}{rll}\text{Number}\text{of}\text{moles}\text{of}{\text{KHCO}}_{\text{3}}& =& 255\text{mL}{\text{CO}}_2\times \frac{\text{1}\text{L}}{\text{1000}\text{mL}}\times \frac{\text{1}\text{mol}{\text{CO}}_{\text{2}}}{\text{22}\text{.4}\text{L}{\text{CO}}_{\text{2}}}\times \frac{\text{2}\text{mol}{\text{KHCO}}_{\text{3}}}{\text{1}\text{mol}{\text{CO}}_{\text{2}}}\\ & =& \text{0}\text{.02276}\text{mol}{\text{KHCO}}_{\text{3}}\\ & \approx & 0.0228\text{mol}{\text{KHCO}}_{\text{3}}\end{array}$

The molar mass of potassium is $39.10\text{g}{\text{mol}}^{-\text{1}}$.

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

The molar mass of hydrogen is $1.01\text{g}{\text{mol}}^{-\text{1}}$.

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

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}\text{of}{\text{KHCO}}_{\text{3}}& =& \left(\begin{array}{l}\text{Molar}\text{mass}\text{of}\text{K}+\text{Molar}\text{mass}\text{of}\text{H}+\\ \text{Molar}\text{mass}\text{of}\text{C}+\left(\text{3}\times \text{Molar}\text{mass}\text{of}\text{O}\right)\end{array}\right)\\ & =& 39.10\text{g}{\text{mol}}^{-\text{1}}+1.01\text{g}{\text{mol}}^{-\text{1}}+12.01\text{g}{\text{mol}}^{-\text{1}}+\left(3\times 16.00\text{g}{\text{mol}}^{-\text{1}}\right)\\ & =& 39.10\text{g}{\text{mol}}^{-\text{1}}+1.01\text{g}{\text{mol}}^{-\text{1}}+12.01\text{g}{\text{mol}}^{-\text{1}}+48.00\text{g}{\text{mol}}^{-\text{1}}\\ & =& 100.12\text{g}{\text{mol}}^{-\text{1}}\end{array}$ The mass of ${\text{KHCO}}_{\text{3}}$ gas is calculated from the relation as shown below.

$\begin{array}{l}1\text{mol}=\text{Molar}\text{mass}\text{of}{\text{KHCO}}_{\text{3}}\\ 1\text{mol}=100.12\text{g}\text{of}{\text{KHCO}}_{\text{3}}\end{array}$

The number of moles of ${\text{KHCO}}_{\text{3}}$ is $\text{0}\text{.0228}\text{mol}$.

The mass for $\text{0}\text{.0228}\text{mol}$ of ${\text{KHCO}}_{\text{3}}$ is calculated as shown below.

$\begin{array}{rll}\text{Mass}\text{of}{\text{KHCO}}_{\text{3}}& =& \text{0}\text{.0228}\text{mol×}\frac{100.12\text{g}{\text{KHCO}}_{\text{3}}}{\text{1}\text{mol}{\text{KHCO}}_{\text{3}}}\\ & =& 2.282\text{g}{\text{KHCO}}_{\text{3}}\end{array}$

Therefore, the mass of ${\text{KHCO}}_{\text{3}}$ is $2.282\text{g}$.

Conclusion

The stoichiometry concept map is shown in Figure 1. The mass of ${\text{KHCO}}_{\text{3}}$ decomposed is $2.282\text{g}$.

Interpretation Introduction

(b)

Interpretation:

The stoichiometry concept map is to be drawn and the mass of the ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ produced is to be stated.

Concept introduction:

Stoichiometry measures the amount of reactant and product consumed or formed in a chemical reaction. It gives the relationship between the reactants and products. Stoichiometry concept maps are used to determine the relation of reactant and product in terms of stoichiometry.

Answer to Problem 26E

The stoichiometry map is shown below.

The mass of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ produced is $1.57\text{g}$.

Explanation of Solution

The stoichiometry concept mass is shown below.

Figure 1

The number of potassium atom and the hydrogen atom are not equal on both sides. A coefficient of $2$ is added in front of ${\text{KHCO}}_3$ to balance the chemical reaction.

The balanced equation is stated below.

${\text{2KHCO}}_3(s)\to {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}(aq)+{\text{H}}_{\text{2}}\text{O}(l)+{\text{CO}}_{\text{2}}\left(g\right)$…(2)

The equation (2) shows that $2$ mole of ${\text{KHCO}}_{\text{3}}$ is consumed to form $1$ mole of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$.

The molar mass of ${\text{KHCO}}_{\text{3}}$ is calculated in part (a) is $100.12\text{g}{\text{mol}}^{-\text{1}}$.

The mass calculated of ${\text{KHCO}}_{\text{3}}$ in part (a) is $2.282\text{g}$.

The number of moles of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ is calculated as shown below.

$\begin{array}{rll}\text{Number}\text{of}\text{moles}\text{of}{\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}& =& 2.282\text{g}{\text{KHCO}}_{\text{3}}\times \frac{\text{1}\text{mol}{\text{KHCO}}_{\text{3}}}{100.12\text{g}{\text{KHCO}}_{\text{3}}}\times \frac{\text{1}\text{mol}{\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}}{\text{2}\text{mol}{\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}}\\ & =& 0.01139\text{mol}\\ & \approx & \text{0}\text{.0114}\text{mol}\end{array}$

The molar mass of potassium is $39.10\text{g}{\text{mol}}^{-\text{1}}$.

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

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

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}\text{of}{\text{K}}_2{\text{CO}}_{\text{3}}& =& \left(\begin{array}{l}\left(\text{2}\times \text{Molar}\text{mass}\text{of}\text{K}\right)+\text{Molar}\text{mass}\text{of}\text{C}\\ +\left(\text{3}\times \text{Molar}\text{mass}\text{of}\text{O}\right)\end{array}\right)\\ & =& \left(2\times 39.10\text{g}{\text{mol}}^{-\text{1}}\right)+12.01\text{g}{\text{mol}}^{-\text{1}}+\left(3\times 16.00\text{g}{\text{mol}}^{-\text{1}}\right)\\ & =& 78.20\text{g}{\text{mol}}^{-\text{1}}+12.01\text{g}{\text{mol}}^{-\text{1}}+48.00\text{g}{\text{mol}}^{-\text{1}}\\ & =& 138.21\text{g}{\text{mol}}^{-\text{1}}\end{array}$

The mass of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ gas is calculated from the relation as shown below.

$\begin{array}{l}1\text{mol}=\text{Molar}\text{mass}\text{of}{\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}\\ 1\text{mol}=138.21\text{g}\text{of}{\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}\end{array}$

The number of moles of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ is $\text{0}\text{.0114}\text{mol}$.

The mass for $\text{0}\text{.0114}\text{mol}$ of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ is calculated as shown below.

$\begin{array}{rll}\text{Mass}\text{of}{\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}& =& \text{0}\text{.0114}\text{mol×}\frac{138.21\text{g}{\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}}{\text{1}\text{mol}{\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}}\\ & =& 1.57\text{g}{\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}\end{array}$

Therefore, the mass of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ is $1.57\text{g}$.

Conclusion

The mass of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ produced after the reaction is $1.57\text{g}$.