Question

mno₂ + 4hcl → mncl₂ + cl₂ + 2h₂o
9.52 g mno₂ reacts with 16.54 g hcl according to the reaction above.
what is the theoretical yield of chlorine in grams?
? g cl₂
be careful to check for the limiting reactant.

Explanation:

Step1: Calculate moles of \( \text{MnO}_2 \)

Molar mass of \( \text{MnO}_2 \) is \( 54.94 + 2\times16.00 = 86.94 \, \text{g/mol} \).
Moles of \( \text{MnO}_2 = \frac{9.52 \, \text{g}}{86.94 \, \text{g/mol}} \approx 0.1095 \, \text{mol} \).

Step2: Calculate moles of \( \text{HCl} \)

Molar mass of \( \text{HCl} \) is \( 1.01 + 35.45 = 36.46 \, \text{g/mol} \).
Moles of \( \text{HCl} = \frac{16.54 \, \text{g}}{36.46 \, \text{g/mol}} \approx 0.4536 \, \text{mol} \).

Step3: Determine limiting reactant

From the reaction \( \text{MnO}_2 + 4\text{HCl}
ightarrow \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O} \), the mole ratio \( \text{MnO}_2 : \text{HCl} = 1:4 \).
Moles of \( \text{HCl} \) required for \( 0.1095 \, \text{mol} \, \text{MnO}_2 \): \( 0.1095 \times 4 = 0.438 \, \text{mol} \).
Available \( \text{HCl} \) is \( 0.4536 \, \text{mol} \), so \( \text{MnO}_2 \) is limiting (since \( 0.438 < 0.4536 \)).

Step4: Calculate moles of \( \text{Cl}_2 \) from limiting reactant

Mole ratio \( \text{MnO}_2 : \text{Cl}_2 = 1:1 \), so moles of \( \text{Cl}_2 = 0.1095 \, \text{mol} \).

Step5: Calculate mass of \( \text{Cl}_2 \)

Molar mass of \( \text{Cl}_2 = 2\times35.45 = 70.90 \, \text{g/mol} \).
Mass of \( \text{Cl}_2 = 0.1095 \, \text{mol} \times 70.90 \, \text{g/mol} \approx 7.77 \, \text{g} \).

Answer:

\( 7.77 \) (or rounded appropriately, e.g., \( 7.76 - 7.78 \) depending on precision)