Chapter 12, Problem 78A

Interpretation Introduction

Interpretation : The percentage of calcium carbonate in the limestone 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 percentage yield is given as

  $\text{Percent}\text{yield =}\frac{\text{actual}\text{yield}}{\text{theoretical}\text{yield}}\times 100\%$   ...... (2)

Answer to Problem 78A

The percentage of calcium carbonate in the limestone is $87.41\%$ .

Explanation of Solution

Mass of limestone is $84.4\text{g}$ .

Mass of calcium chloride formed is $81.8\text{g}$ .

The given chemical equation is as follows:

  ${\text{CaCO}}_{\text{3}}+2\text{HCl} \to {\text{CaCl}}_2+{\text{H}}_2\text{O}+{\text{CO}}_{\text{2}}$

Molar mass of $\text{C}$ is $12.01\text{g/mol}$ .

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

Molar mass of $\text{Ca}$ is $\text{40}\text{.08}\text{g/mol}$ .

Molar mass of $\text{Cl}$ is $35.45\text{g/mol}$ .

Mass of ${\text{CaCO}}_{\text{3}}$ is given as follows:

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

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

  $\begin{array}{c}\text{Mass of}{\text{CaCl}}_2=\text{M}\left(\text{Ca}\right)\text{+2M}\left(\text{Cl}\right)\\ =\left(40.08+\left(2\times 35.45\right)\right)\text{g/mol}\\ \text{= 110}\text{.98}\text{g/mol}\end{array}$

To calculate the moles of ${\text{CaCl}}_2$ , substitute the values in equation (1).

  $\begin{array}{c}{\text{n}}_{{\text{CaCl}}_2}\text{=}\frac{{\text{m}}_{{\text{CaCl}}_2}}{\text{M}\left({\text{CaCl}}_2\right)}\\ =\frac{81.8\cancel{\text{g}}}{110.98\cancel{\text{g}}\text{/mol}}\\ =0.7371\text{mol}\end{array}$

The moles of ${\text{CaCl}}_2$ is equal to the moles of ${\text{CaCO}}_{\text{3}}$ .

So, the moles of ${\text{CaCO}}_{\text{3}}$ is $0.7371\text{mol}$ .

To calculate the mass of ${\text{CaCO}}_{\text{3}}$ , substitute the values in equation (1).

  $\begin{array}{c}{\text{m}}_{{\text{CaCO}}_{\text{3}}}={\text{n}}_{{\text{CaCO}}_{\text{3}}}\times \text{M}\left({\text{CaCO}}_{\text{3}}\right)\\ =0.7371\cancel{\text{mol}}\times 100.09\text{g/}\cancel{\text{mol}}\\ =73.77\text{g}\end{array}$

The theoretical mass of limestone is $84.4\text{g}$ .

To calculate the actual mass of ${\text{CaCO}}_{\text{3}}$ , substitute the values in equation (2).

  $\begin{array}{c}\text{Percent}\text{yield}\text{=}\frac{\text{actual}\text{yield}}{\text{theoretical}\text{yield}}\times 100\%\\ =\frac{73.77\cancel{\text{g}}}{84.4\cancel{\text{g}}}\times 100\%\\ =87.41\%\end{array}$