Chapter 12, Problem 77A

Interpretation Introduction

Interpretation : The density of the carbon dioxide is to be calculated.

Concept Introduction : The coefficients of a balanced chemical equation that are transformed into moles are used to calculate the mole ratio, which is a conversion factor.

When converting between a certain number of moles of one reactant or product and moles of another reactant or product, mole ratios are utilized.

The formula to calculate the number of moles is given as

  $\text{n}\text{=}\frac{\text{given}\text{mass}}{\text{molar mass}}$   ...... (1)

The formula to calculate the density is given as

  $\text{density}\text{=}\frac{\text{mass}}{\text{volume}}$   ...... (2)

Answer to Problem 77A

The density of ${\text{CO}}_2$ is $1.965\text{g/L}$ .

Explanation of Solution

The sample of ${\text{CaCO}}_3$ is $1004.0\text{g}$ .

The purity of ${\text{CaCO}}_3$ is $95\%$ .

The volume of ${\text{CO}}_2$ is $225\text{L}$ .

The chemical equation between calcium carbonate and hydrochloric acid is given as

  ${\text{CaCO}}_{\text{3}}\text{+2HCl}\to {\text{CaCl}}_{\text{2}}{\text{+CO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}$

Since the purity of ${\text{CaCO}}_3$ is $95\%$ , the mass of ${\text{CaCO}}_3$ is given as

  $\begin{array}{c}{\text{m}}_{{\text{CaCO}}_{\text{3}}}=0.95\times 1004.0\text{g}\\ \text{= 953}\text{.8}\text{g}\end{array}$

Molar mass of $\text{C}$ is $\text{12}\text{.00}\text{g/mol}$ .

Molar mass of $\text{H}$ is $\text{1}\text{.008}\text{g/mol}$ .

Molar mass of $\text{O}$ is $\text{16}\text{.00}\text{g/mol}$ .

Molar mass of $\text{Ca}$ is $40.08\text{g/mol}$ .

Mass of ${\text{CaCO}}_3$ is given as follows:

  $\begin{array}{c}\text{Mass of}{\text{CaCO}}_3=\text{M}\left(\text{Ca}\right)\text{+M}\left(\text{C}\right)+3\text{M}\left(\text{O}\right)\\ =\left(40.08+12.01+\left(3\times 16.00\right)\right)\text{g/mol}\\ \text{= 100}\text{.09}\text{g/mol}\end{array}$

Mass of ${\text{CO}}_2$ is given as follows:

  $\begin{array}{c}\text{Mass of}{\text{CO}}_2=\text{M}\left(\text{C}\right)\text{+2M}\left(\text{O}\right)\\ =\left(12.01+\left(2\times 16.00\right)\right)\text{g/mol}\\ \text{= 44}\text{.01}\text{g/mol}\end{array}$

To calculate the number of moles of ${\text{CaCO}}_3$, substitute the values in equation (1):

  $\begin{array}{c}{\text{n}}_{{\text{CaCO}}_3}\text{=}\frac{{\text{m}}_{{\text{CaCO}}_3}}{{\text{M}}_{{\text{CaCO}}_3}}\\ =\frac{\text{953}\text{.8}\cancel{\text{g}}}{100.04\cancel{\text{g}}\text{/mol}}\\ =9.53\text{mol}\end{array}$

The number of moles of ${\text{CaCO}}_3$ is equal to the number of moles of ${\text{CO}}_2$ according to the chemical equation.

So, the number of moles of ${\text{CO}}_2$ is $9.53\text{mol}$ .

To calculate the mass of ${\text{CO}}_2$, substitute the values in equation (1):

  $\begin{array}{c}{\text{m}}_{{\text{CO}}_2}={\text{n}}_{{\text{CO}}_2}\times {\text{M}}_{{\text{CO}}_2}\\ =9.53\cancel{\text{mol}}\times 44.01\text{g/}\cancel{\text{mol}}\\ =419.415\text{g}\end{array}$

At STP means Standard Temperature $\left(0^{\circ }\text{C}\right)$ and pressure $\left(1\text{atm}\text{=}\text{101325}\text{Pa}\right)$ .

The molar volume is $22.4{\text{dm}}^3{\text{mol}}^{-1}$ .

The formula of volume is given as

  ${\text{V=V}}_{\text{m}}\text{×n}$

To calculate the volume of ${\text{CO}}_2$, substitute the values in the above formula:

  $\begin{array}{c}{\text{V=V}}_{\text{m}}\text{×n}\\ \text{=}22.4{\text{dm}}^3\cancel{{\text{mol}}^{-1}}\times 9.53\cancel{\text{mol}}\\ =213.472{\text{dm}}^{\text{3}}\end{array}$

To calculate the density of ${\text{CO}}_2$, substitute the values in equation (2):

  $\begin{array}{c}\text{density}\text{=}\frac{\text{mass}}{\text{volume}}\\ =\frac{419.415\text{g}}{213.472{\text{dm}}^3}\\ =1.965{\text{g/dm}}^3\left(\because 1{\text{dm}}^{\text{3}}=1\text{ L}\right)\\ =1.965\text{g/L}\end{array}$