Question

consider the following balanced reaction. if you begin with 25.00 grams of h₂ and 10.00 grams of cl₂, which is the limiting reactant and how much hcl will be formed?
h₂ + cl₂ → 2 hcl
○ h₂; 24.75 g of hcl
○ cl₂; 5.142 g of hcl
○ h₂; 361.0 g of hcl
○ cl₂; 10.28 g of hcl

Explanation:

Step1: Find moles of \( H_2 \) and \( Cl_2 \)

Molar mass of \( H_2 = 2 \, g/mol \), moles of \( H_2 = \frac{25.00 \, g}{2 \, g/mol} = 12.5 \, mol \).
Molar mass of \( Cl_2 = 71 \, g/mol \), moles of \( Cl_2 = \frac{10.00 \, g}{71 \, g/mol} \approx 0.1408 \, mol \).

Step2: Determine limiting reactant

From reaction \( H_2 + Cl_2
ightarrow 2HCl \), mole ratio \( H_2:Cl_2 = 1:1 \).
\( Cl_2 \) has fewer moles, so \( Cl_2 \) is limiting.

Step3: Calculate moles of \( HCl \) from \( Cl_2 \)

Mole ratio \( Cl_2:HCl = 1:2 \), so moles of \( HCl = 2 \times 0.1408 \, mol \approx 0.2816 \, mol \).

Step4: Calculate mass of \( HCl \)

Molar mass of \( HCl = 36.46 \, g/mol \), mass of \( HCl = 0.2816 \, mol \times 36.46 \, g/mol \approx 10.28 \, g \).

Answer:

Cl₂; 10.28 g of HCl