Chapter 11, Problem 96A

Interpretation Introduction

Interpretation:

The theoretical yield and percent yieldof the production of P₄ needs to be determined.

Concept introduction:

Phosphorus (P₄) is commercially produced by heating a mixture of calcium phosphate, sand and coke in electric furnace.

Answer to Problem 96A

Theoretical yield of P₄ = 49.92 g

Percent yield = 90.14%

Explanation of Solution

Step 1: write balance chemical equation

$\begin{array}{l}2Ca3(PO4)2(s)+6SiO2(s)\to 6CaSiO3(l)+P4O10(g)\\ 250g400g\\ P4O10(g)+10C(s)\to P4(g)+10CO(g)\\ ?\end{array}$

Step 2: convert mass of reactant to moles

$\begin{array}{c}MolarmassofCa3(PO4)2=310.18g/mole\\ Moles=\frac{Mass}{Molarmass}\\ MolesofCa3(PO4)2=\frac{250}{310.18}=0.806moles\\ MolarmassofSiO2=60.084g/mole\\ Moles=\frac{Mass}{Molarmass}\\ MolesofSiO2=\frac{400}{60.084}=6.657moles\end{array}$

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

2 moles of Ca₃(PO₄)₂ produce = 1 moles of P₄ 0.806 moles of CO produce = 0.403 moles of P₄ 6 moles ofSiO₂ produce = 1 moles of P₄ 6.657 moles of SiO₂ produce = $\frac{1}{6}\times 6.657$ moles of P₄ = 1.1095 moles of P₄

Ca₃(PO₄)₂ produce a smaller number of moles of P₄. So, acts as limiting reactant.

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

$\begin{array}{c}Molar\text{}massofP4=123.89g/mole\\ Mass=Molarmass\times Moles\\ MassofP4(Theoreticalyield)=123.89\times 0.403=49.92g\end{array}$

Actual yield = 45 g of CH₃OH

$\begin{array}{c}PercentYield=\frac{Actualyield}{Theoreticalyield}\times 100\\ =\frac{45}{49.92}\times 100=90.14\%\end{array}$