Chapter 9, Problem 68A

Interpretation Introduction

Interpretation:The limiting reactant and mass of leftover reactant needs to be determined for the given reaction when 50.0 g of $\text{CaO}$ reacts with 50.0 g of carbon.

Concept introduction: The reagent in reaction that controls the amount of product formed is termed as limiting reagent. After completion of chemical reaction, the limiting reagent is fully utilized.

Answer to Problem 68A

  $\text{CaO}$ is the limiting reactant and the mass of left-over reactant is $23.18\text{ g}$ of $\text{C}$ .

Explanation of Solution

Given:

The balanced reaction is:

  $\text{2CaO}\left(s\right)+\text{5C}\left(s\right)\to {\text{2CaC}}_{\text{2}}\left(s\right)+{\text{CO}}_2\left(g\right)$

The mass of $\text{CaO}$ and $\text{C}$ is 50.0 g.

The molar mass of $\text{CaO}$ and $\text{C}$ is 56.08 g/mol and 12.01 g/mol respectively. The number of moles of each is calculated using formula:

  $\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}$

So,

  $\begin{array}{l}{\text{n}}_{\text{CaO}}\text{= }\frac{\text{50}\text{.0 g}}{\text{56}\text{.08 g/mol}}\\ {\text{n}}_{\text{CaO}}\text{= 0}\text{.892 mol}\end{array}$

  $\begin{array}{l}{\text{n}}_{\text{C}}\text{= }\frac{\text{50}\text{.0 g}}{\text{12}\text{.01 g/mol}}\\ {\text{n}}_{\text{C}}\text{= 4}\text{.16 mol}\end{array}$

In the given balanced reaction, 2.0 mol of $\text{CaO}$ reacts with 5.0 mol of $\text{C}$ . So, the number of moles of $\text{CaO}$ required to react with $\text{4}\text{.16 mol}$ of $\text{C}$ is:

  $4.16\text{ mol of C×}\frac{\text{2 mol of CaO}}{\text{5 mol of C}}\text{ = }1.66\text{ mol of CaO}$

Since, the required moles of $\text{CaO}$ is $1.66\text{ mol}$ and available amount is $\text{0}\text{.892 mol}$ so, $\text{CaO}$ is the limiting reactant and the C is present in excess.

The mass of leftover reactant is calculated as:

In the given balanced reaction, 2.0 mol of $\text{CaO}$ reacts with 5.0 mol of $\text{C}$ . So, the number of moles of $\text{C}$ required to react with $\text{0}\text{.892 mol}$ of $\text{CaO}$ (limiting reactant) is:

  $\text{0}\text{.892 mol of CaO×}\frac{\text{5 mol of C}}{\text{2 mol of CaO}}\text{ = }2.23\text{ mol of C}$

Thus, the leftover mole of C is:

  $\text{4}\text{.16 mol - }2.23\text{ mol = 1}\text{.93 mol}$

Since, the molar mass of C is 12.011 g/mol so:

  $\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}$

  $\begin{array}{l}\text{1}\text{.93 mol}=\frac{\text{Mass}}{\text{12}\text{.011 g/mol}}\\ \text{Mass = 1}\text{.93 mol}\times \text{12}\text{.011 g/mol}\\ \text{Mass = }23.18\text{ g}\end{array}$

Hence, the mass of left-over reactant is $23.18\text{ g}$ of $\text{C}$ .