Chapter 11, Problem 92A

a)

Interpretation Introduction

Interpretation : On heating Calcium carbonate, it decomposes to Calcium oxide and Carbon dioxide. If 235g of CaCO₃ is given then the theoretical yield needs to be calculated.

Concept introduction : Theoretical yield is the maximum amount of product that can be formed on completion of reaction.

Answer to Problem 92A

Theoretical yield = 103.3 g of CO_2.

Explanation of Solution

Step 1: write balance chemical equation

$\begin{array}{l}CaCO3(s)\to CaO(s)+CO2(g)\\ 235g?\end{array}$

Step 2: convert mass of reactant to moles

$\begin{array}{c}MolarmassofCaCO3=100.089g/mole\\ Moles=\frac{Mass}{Molarmass}\\ MolesofCaCO3=\frac{235}{100.089}=2.34moles\end{array}$

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

1 moles of CaCO₃ produce = 1 moles of CO₂ 2.34 moles of CaCO₃ produce = 2.34 X 1 moles of CO₂ = 2.34moles of CO₂

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

$\begin{array}{c}Molar\text{}massofCO2=44g/mole\\ Mass=Molarmass\times Moles\\ MassofCO2(Theoreticalyield)=44\times 2.34=103.30g\end{array}$

b)

Interpretation Introduction

Interpretation:

If 97.5g of CO₂ was produced, the percent yield needs to be calculated.

Concept introduction:

The percent yield can be calculated as follows:

$PercentYield=\frac{Actualyield}{Theoreticalyield}\times 100$

Answer to Problem 92A

Percent yield = 94.38%

Explanation of Solution

Actual yield = 97.5 g (given)

$\begin{array}{c}PercentYield=\frac{Actualyield}{Theoreticalyield}\times 100\\ =\frac{97.5}{103.3}\times 100=94.38\%\end{array}$