Question

missed this? watch kcv limiting reactant, theoretical yield, and percent yield. we finding limiting reactant and theoretical yield. read section 8.6. you can click on the review link to access the section in your etext. 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 determine the limiting reactant for this reaction. express your answer as a chemical formula. part b complete previous part(s) part c complete previous part(s) provide feedback

Explanation:

Step1: Calculate moles of reactants

Molar mass of \(HCl\): \(M_{HCl}=1 + 35.5=36.5\space g/mol\). Moles of \(HCl\), \(n_{HCl}=\frac{63.1\space g}{36.5\space g/mol}\approx1.73\space mol\)
Molar mass of \(O_{2}\): \(M_{O_{2}} = 32\space g/mol\). Moles of \(O_{2}\), \(n_{O_{2}}=\frac{17.2\space g}{32\space g/mol}= 0.5375\space mol\)

Step2: Use mole ratio from balanced equation

From \(4HCl(g)+O_{2}(g)\to2H_{2}O(l)+2Cl_{2}(g)\), mole ratio of \(HCl\) to \(O_{2}\) is \(4:1\)
If \(O_{2}\) is limiting, moles of \(HCl\) required \(=4\times n_{O_{2}}=4\times0.5375 = 2.15\space mol\)
But we have \(n_{HCl}=1.73\space mol<2.15\space mol\)

Step3: Determine limiting reactant

Since the available moles of \(HCl\) is less than the moles of \(HCl\) required if \(O_{2}\) were to react completely, \(HCl\) is the limiting reactant.

Answer:

\(HCl\)