Question

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me finding limiting reactant and theoretical yield. read section 8.6. you can click on
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consider the reaction between hcl and o₂.
4hcl(g) + o₂(g) → 2h₂o(l) + 2cl₂(g)
when 63.1 g of hcl is allowed to react with 17.2 g of o₂, 48.3 g of cl₂ is collected.
part a
part b
part c
determine the percent yield of this reaction.
percent yield =

Explanation:

Step1: Calculate moles of reactants

• Molar mass of \(HCl\): \(M_{HCl}=1.01 + 35.45=36.46\space g/mol\)
• Moles of \(HCl\): \(n_{HCl}=\frac{63.1\space g}{36.46\space g/mol}\approx1.73\space mol\)
• Molar mass of \(O_{2}\): \(M_{O_{2}} = 32.00\space g/mol\)
• Moles of \(O_{2}\): \(n_{O_{2}}=\frac{17.2\space g}{32.00\space g/mol}=0.5375\space mol\)

Step2: Determine limiting reactant

From the balanced equation \(4HCl(g)+O_{2}(g)\to2H_{2}O(l) + 2Cl_{2}(g)\), the mole ratio of \(HCl\) to \(O_{2}\) is \(4:1\).

• For \(n_{O_{2}} = 0.5375\space mol\), moles of \(HCl\) required \(=4\times0.5375 = 2.15\space mol\)
• Since \(n_{HCl}=1.73\space mol<2.15\space mol\), \(HCl\) is the limiting reactant.

Step3: Calculate theoretical yield of \(Cl_{2}\)

From the balanced equation, mole ratio of \(HCl\) to \(Cl_{2}\) is \(4:2 = 2:1\)

• Moles of \(Cl_{2}\) from \(HCl\): \(n_{Cl_{2}}=\frac{1.73\space mol}{2}=0.865\space mol\)
• Molar mass of \(Cl_{2}\): \(M_{Cl_{2}}=70.90\space g/mol\)
• Theoretical yield of \(Cl_{2}\): \(m_{theoretical}=0.865\space mol\times70.90\space g/mol\approx61.3\space g\)

Step4: Calculate percent yield

Percent yield formula: \(\text{Percent yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times100\%\)

• Given actual yield \(m_{actual} = 48.3\space g\)
• \(\text{Percent yield}=\frac{48.3\space g}{61.3\space g}\times100\%\approx78.8\%\)

Answer:

\(78.8\%\)