Chapter 11, Problem 97A

Interpretation Introduction

Interpretation:

The theoretical yield and percent yield of the production of Cl₂ needs to be determined.

Concept introduction:

Chlorine produced by the reaction of hydrochloric acid and manganese (IV) oxide.

Answer to Problem 97A

Theoretical yield of Cl₂ = 24.28 g

Percent yield = 82.37%

Explanation of Solution

Step 1: write balance chemical equation

$\begin{array}{l}MnO2+4HCl\to MnCl2+Cl2+2H2O\\ 86g50g20g\end{array}$

Step 2: convert mass of reactant to moles

$\begin{array}{c}MolarmassofMnO2=86.93g/mole\\ Moles=\frac{Mass}{Molarmass}\\ MolesofMnO2=\frac{86}{86.93}=0.989moles\\ MolarmassofHCl=36.46g/mole\\ Moles=\frac{Mass}{Molarmass}\\ MolesofHCl=\frac{50}{36.46}=1.37moles\\ MolarmassofCl2=70.90g/mole\\ Moles=\frac{Mass}{Molarmass}\\ MolesofCl2=\frac{20}{70.90}=0.28moles\end{array}$

Step 3: Convert moles of reactant to moles of product using mole ratio

1 moles ofMnO₂ produce = 1 moles of Cl₂ 0.989 moles of CO produce = 0.989 moles of Cl₂ 4 moles of HCl produce = 1 moles of Cl₂ 1.37 moles of HCl produce = $\frac{1}{4}\times 1.37$ moles of Cl₂ = 0.3425 moles of Cl₂

HCl produce a smaller number of moles of Cl₂. So, acts as limiting reactant.

Step 4: Convert moles of product to mass (theoretical yield)

$\begin{array}{c}Molar\text{}massofCl2=70.9g/mole\\ Mass=Molarmass\times Moles\\ MassofCl2(Theoreticalyield)=70.9\times 0.403=24.28g\end{array}$

Actual yield = 20 g of Cl₂

$\begin{array}{c}PercentYield=\frac{Actualyield}{Theoreticalyield}\times 100\\ =\frac{20}{24.28}\times 100=82.37\%\end{array}$