Chapter 9, Problem 64A

Interpretation Introduction

Interpretation:

When 50.0 g sample of CaO is reacted with 50.0 g of carbon. The mass of left -over reactant and limiting reactant should be predicted.

Concept Introduction: The reaction in which the reactant is totally consumed when the reaction is completed is known as limiting reactant.

Answer to Problem 64A

Mass of left- over reactant is 2.23 mol

CaO is limiting reactant

Explanation of Solution

The calculation of balanced chemical reaction is shown below:

  $\text{2CaO}\left(\text{s}\right)\text{+}\text{5C}\left(\text{s}\right)\to {\text{2CaC}}_{\text{2}}\left(\text{s}\right)\text{+}{\text{CO}}_{\text{2}}\left(\text{g}\right)$

Given information is as follows:

Mass of CaO = 50 g

molar mass of CaO = 56 g

The calculation of moles is shown below:

  $\begin{array}{c}\text{moles of CaO = }\frac{\text{ mass}\text{of}\text{CaO}}{\text{molar}\text{mass}\text{of}\text{CaO}}\\ \text{= }\frac{\text{50}\text{g}}{\text{56}{\text{gmol}}^{\text{-1}}}\\ \text{ = 0}\text{.8929}\text{mol}\end{array}$

Also,

  $\begin{array}{c}\text{moles of C = }\frac{\text{mass}\text{of}\text{C }}{\text{molar}\text{mass}\text{of}\text{C}}\\ \text{ = }\frac{\text{50}\text{g }}{\text{12}{\text{gmol}}^{\text{-1}}}\\ \text{ = 4}\text{.17}\text{mol}\end{array}$

According to the balanced reaction:

  $\begin{array}{l}\text{2 moles of CaO reacts = 5 moles of C}\hfill \\ \text{ 1 mole CaO reacts = }\frac{\text{5}}{2}\text{ moles}\text{of}\text{C}\hfill \\ \begin{array}{l}\text{ 0}\text{.8929 moles of CaO = 2}\text{.5×0}\text{.8929}\text{moles}\text{of}\text{C}\\ \text{= 2}\text{.23 moles of C}\end{array}\hfill \end{array}$

But 4.17 moles of C are initially present, so C is present in excess. Therefore; CaO is the limiting reactant.

Conclusion

Therefore,

Mass of left- over reactant is 2.23 mol

CaO is limiting reactant.