Chapter 9, Problem 48A

Interpretation Introduction

Interpretation:

155 g of calcium carbonate is treated with 250 g of anhydrous hydrogen chloride and only 142 g of CaCl₂ is obtained, the percentage yield of the compound should be predicted.

Concept Introduction:

The reaction in which the reactant is totally consumed when the reaction is completed isknown as limiting reactant.

The ratio of actual yield to the theoretical yield multiplied by 100 is known as percent yield.

Answer to Problem 48A

The percentage yield is 76.49%

Explanation of Solution

The given reaction is shown below:

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

The calculation of moles is shown below:

  $\text{Moles of HCl}\text{=}\frac{\text{mass}\text{of}\text{HCl}}{\text{molar}\text{mass}\text{of}\text{HCl}}$

  $\text{=}\frac{\text{250}\text{g}}{\text{36}\text{.5}{\text{gmol}}^{\text{-1}}}$

  $\text{ =}\text{6}\text{.84}\text{mol}$

  $\begin{array}{l}{\text{Moles of CaCO}}_{\text{3}}\text{=}\frac{\text{mass}\text{of}{\text{CaCO}}_{\text{3}}}{\text{molar}\text{mass}\text{of}{\text{CaCO}}_{\text{3}}}\\ \text{=}\frac{\text{405}\text{g}}{\text{100}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{4}\text{.05}\text{mol}\end{array}$

  $\begin{array}{l}{\text{Moles of CaCl}}_{\text{2}}\text{ produced actually =}\frac{\text{mass}\text{of}{\text{CaCl}}_{\text{2}}}{\text{molar}\text{mass}\text{of}{\text{CaCl}}_{\text{2}}}\\ \text{= }\frac{\text{242}\text{g}}{\text{111}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{2}\text{.18}\text{mol}\end{array}$

According to the given reaction 2 mole of HCl reacts with 1 mole of CaCO₃ and produced 1

mol of CaCl₂. Therefore, 4.05 moles of CaCO₃ would react with 2× 4.05 = 8.1 moles of HCl

But there are 6.84 moles of HCl that means HClwill get totally consumed in the reaction and thus, will act as a limiting reactant.

  $\begin{array}{l}{\text{Moles of CaCl}}_{\text{2}}\text{produced theoretical=}\frac{\text{6}\text{.84}}{\text{2}}\\ \text{=}\text{3}\text{.42}\end{array}$

Moles of H₂O produced =3.42

According to the given reaction;

6 moles of H₂O =1 mole of CaCl₂

  $\begin{array}{l}\text{3}{\text{.42 moles of H}}_{\text{2}}\text{O =}\frac{\text{3}\text{.42}}{\text{6}}{\text{moles of CaCl}}_{\text{2}}\\ \text{=}\text{0}{\text{.57 moles of CaCl}}_{\text{2}}\end{array}$

actual theoretical yield =3.42-0.57

  = 2.85

The calculation of percentage yield is shown below:

  $\begin{array}{l}\text{ percentage yield =}\frac{\text{actual}\text{yield}}{\text{theoretical}\text{yield}}\text{×}\text{100}\\ \text{= }\left(\frac{\text{2}\text{.18}}{\text{2}\text{.85}}\right)\text{×100}\\ \text{=76}\text{.49\%}\end{array}$

Conclusion

Thus, the percentage yield is 76.49%