Chapter 12.3, Problem 32SP

Interpretation Introduction

Interpretation: The percent yield of the given reaction is to be stated.

Concept Introduction: According to the law of conservation of mass, the mass can neither be created nor be destroyed. Every reaction must be balanced. With the help of the balanced equation, the limiting reagent can be found and in accordance with it, the amount of product can be calculated.

Answer to Problem 32SP

The percent yield is 83.8%.

Explanation of Solution

Given:

• 50.0 g of silicon dioxide
• Excess of carbon
• Theoretical yield of silicon carbide is 27.9 g

The reaction is given as follows:

  ${\text{SiO}}_{\text{2}}\left(s\right)+3\text{C}\left(s\right)\stackrel{\Delta }{\to }\text{SiC}\left(s\right)+2\text{CO}\left(g\right)$

The mol of silicon dioxide is calculated as:

  $\begin{array}{c}\text{Mole}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{50.0\text{g}}{60\text{g/mol}}\\ =0.83\text{mol}\end{array}$

According to the balanced equation,

  $\begin{array}{c}\text{1}\text{mol}\text{of}{\text{SiO}}_{\text{2}}\to \text{1}\text{mol}\text{of}\text{SiC}\\ \text{0}\text{.83}\text{mol}\text{of}{\text{SiO}}_{\text{2}}\to \text{0}\text{.83}\text{mol}\text{of}\text{SiC}\end{array}$

The mass of silicon carbide is calculated as:

  $\begin{array}{c}\text{Mass}=\text{Mole}\times \text{Molar}\text{mass}\\ =0.83\text{mol}\times 40.11\text{g/mol}\\ =\text{33}\text{.3}\text{g}\end{array}$

The theoretical yield of silicon carbide is 33.3 g and the actual yield is given as 27.9 g. The percent yield is calculated as:

  $\begin{array}{c}\text{\%}\text{yield}=\frac{\text{Actual}\text{yield}}{\text{Theoretical}\text{yield}}\times 100\\ =\frac{27.9\text{g}}{\text{33}\text{.3}\text{g}}\times 100\\ =83.8\%\end{array}$

Hence, the percent yield is 83.8%.