Question

missed this? watch kcv: limiting reactant, theoretical yield, and percent yield we: finding limiting reactant and theoretical yield read section 3.6. 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, 40.0 g of cl₂ is collected. part b determine the theoretical yield of cl₂ for this reaction. submit request answer g part c complete previous part(s) provide feedback

Explanation:

Step1: Calculate moles of HCl

Molar mass of HCl is $36.46\ \text{g/mol}$. Moles of HCl = $\frac{63.1\ \text{g}}{36.46\ \text{g/mol}} \approx 1.731\ \text{mol}$.

Step2: Use stoichiometry from reaction

The reaction is $4\text{HCl}(g) + \text{O}_2(g)
ightarrow 2\text{H}_2\text{O}(l) + 2\text{Cl}_2(g)$. From stoichiometry, 4 moles of HCl produce 2 moles of $\text{Cl}_2$. So moles of $\text{Cl}_2$ from HCl = $\frac{2}{4} \times 1.731\ \text{mol} = 0.8655\ \text{mol}$.

Step3: Calculate theoretical mass of $\text{Cl}_2$

Molar mass of $\text{Cl}_2$ is $70.90\ \text{g/mol}$. Theoretical mass = $0.8655\ \text{mol} \times 70.90\ \text{g/mol} \approx 61.4\ \text{g}$.

Answer:

$\approx 61.4\ \text{g}$