Question

missed this? watch kcv: limiting reactant, theoretical yield, and percent yield, iwe: finding limiting reactant, theoretical yield, iread 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, 69.9 g of cl₂ is collected. part a determine the limiting reactant for this reaction. express your answer as a chemical formula. a chemical reaction does not occur for this question. submit reattempt answer part b complete previous part(s) part c complete previous part(s)

Explanation:

Step1: Calculate moles of HCl

Molar mass of HCl is \(36.46\space g/mol\). Moles of HCl \(n_{HCl}=\frac{63.1\space g}{36.46\space g/mol}\approx1.73\space mol\).

Step2: Calculate moles of \(O_2\)

Molar mass of \(O_2\) is \(32.00\space g/mol\). Moles of \(O_2\) \(n_{O_2}=\frac{17.2\space g}{32.00\space g/mol}\approx0.5375\space mol\).

Step3: Determine mole ratio from reaction

Reaction: \(4HCl(g)+O_2(g)
ightarrow2H_2O(l)+2Cl_2(g)\). Mole ratio of \(HCl:O_2 = 4:1\).

Step4: Find required moles of \(O_2\) for HCl

For \(1.73\space mol\) of HCl, required \(O_2\) moles \(n_{O_2}^{req}=\frac{1.73}{4}=0.4325\space mol\).

Step5: Compare available and required \(O_2\)

Available \(O_2 = 0.5375\space mol\), required \(O_2 = 0.4325\space mol\). Since \(n_{HCl}\) requires less \(O_2\) than available, HCl is limiting? Wait, no: Wait, mole ratio HCl to O2 is 4:1. Let's check for O2: required HCl for \(0.5375\space mol\) O2 is \(4\times0.5375 = 2.15\space mol\). But we have only \(1.73\space mol\) HCl. So HCl is limiting because we don't have enough HCl to react with all O2. Wait, my mistake earlier. Let's recalculate:

From reaction, 4 moles HCl react with 1 mole O2.

Moles of HCl: \(63.1\space g\div36.46\space g/mol\approx1.73\space mol\)

Moles of O2: \(17.2\space g\div32\space g/mol = 0.5375\space mol\)

For HCl to react completely, O2 needed: \(1.73\div4 = 0.4325\space mol\) (which is less than 0.5375, so O2 is in excess). For O2 to react completely, HCl needed: \(0.5375\times4 = 2.15\space mol\) (but we have only 1.73 mol HCl, so HCl is limiting.

Answer:

HCl