Chapter 12, Problem 61A

a.

Interpretation Introduction

Interpretation: To calculate the percent yield for the reaction.

Concept Introduction: Percent yield can be calculated by dividing the actual yield to the theoretical yield multiplied by 100.

Answer to Problem 61A

The percent yield of this reaction is $95.2\text{\%}$ .

Explanation of Solution

The given chemical reaction is depicted as follows:

  ${\text{SiO}}_2\left(s\right)+3\text{C}\left(s\right)\to \text{SiC}\left(s\right)+2\text{CO}\left(g\right)$

When one mole of silicon dioxide reacts with three moles of carbon, it yields silicon carbide and two moles of carbon monoxide.

The molar mass of ${\text{SiO}}_2$ is $\text{60}{\text{.08 g mol}}^{-1}$ .

The molar mass of $\text{SiC}$ is $40.11{\text{ g mol}}^{-1}$ .

Calculation for grams of silicon carbide is as follows:

  $\text{50}\text{.0 }\cancel{{\text{g SiO}}_2}\times \frac{1\cancel{{\text{mol SiO}}_2}}{60.08\cancel{{\text{g SiO}}_2}}\times \frac{1\cancel{\text{mol SiC}}}{1\cancel{{\text{mol SiO}}_2}}\times \frac{40.11\text{ g SiC}}{1\cancel{\text{mol SiC}}}=33.38\text{ g SiC}$

The calculation for percent yield is as follows:

  $\begin{array}{c}\text{\%yield}=\frac{\text{ 32}\text{.2 }\cancel{{\text{g Fe}}_2{\text{O}}_3}}{\text{33}\text{.38 }\cancel{{\text{g Fe}}_2{\text{O}}_3}}\text{×100\%}\\ =95.2\text{\%}\end{array}$

b.

Interpretation Introduction

Interpretation: To calculate the amount of carbon monoxide in grams.

Concept Introduction: A mole is $6.02\times {10}^{23}$ atoms/molecules. The SI unit for the amount of substance is a mole and it is represented by “mol”.

The coefficients are the numbers that are written in front of the chemical formula/symbol in order to balance a chemical reaction.

Answer to Problem 61A

The amount of carbon monoxide in grams is $46.6\text{ g}$ .

Explanation of Solution

The given chemical reaction is depicted as follows:

  ${\text{SiO}}_2\left(s\right)+3\text{C}\left(s\right)\to \text{SiC}\left(s\right)+2\text{CO}\left(g\right)$

When one mole of silicon dioxide reacts with three moles of carbon, it yields silicon carbide and two moles of carbon monoxide.

The amount of carbon monoxide can be calculated as follows:

  $\text{50}\text{.0 }\cancel{{\text{g SiO}}_2}\times \frac{1\cancel{{\text{mol SiO}}_2}}{60.08\cancel{{\text{g SiO}}_2}}\times \frac{2\cancel{\text{mol CO}}}{1\cancel{{\text{mol SiO}}_2}}\times \frac{\text{ 28}\text{.01 g CO}}{1\cancel{\text{mol CO}}}=46.6\text{ g CO}$