Question

For reaction

$MnO_2 + 4HCl \longrightarrow MnCl_2 + 2H_2O + Cl_2(g)$

261 gm $MnO_2$ mixed with 448 litre of HCl gas at $273^\circ C$ & 1 atm pressure to produce pro $[N_A = 6 \times 10^{23}$; Atomic mass of Mn = 55, Cl = 35.5]

Select correct statement(s).

• A. $MnO_2$ is limiting reagent
• B. Chlorine gas produced contains $15 \times 10^{23}$ molecules.
• C. Moles of excess reactant left is 0.5 moles
• D. If % yield of reaction is 50%, then mass of $MnCl_2$ obtained will be 315.

Answer 1

Answer: (B), (C)

B, C

Answer 2

The problem involves stoichiometry, limiting reagent calculation, and yield determination for the given reaction:

$MnO_2 + 4HCl \longrightarrow MnCl_2 + 2H_2O + Cl_2(g)$

1. Calculate molar masses of reactants and products:

• Molar mass of $MnO_2 = 55 + 2 \times 16 = 87 \text{ g/mol}$
• Molar mass of $HCl = 1 + 35.5 = 36.5 \text{ g/mol}$
• Molar mass of $Cl_2 = 2 \times 35.5 = 71 \text{ g/mol}$
• Molar mass of $MnCl_2 = 55 + 2 \times 35.5 = 55 + 71 = 126 \text{ g/mol}$

2. Calculate initial moles of reactants:

• Moles of $MnO_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{261 \text{ gm}}{87 \text{ gm/mol}} = 3 \text{ mol}$
• Moles of HCl gas (using Ideal Gas Law, PV=nRT):

Given: P = 1 atm, V = 448 L, T = $273^\circ C = 273 + 273 = 546$ K, R = 0.0821 L atm/mol K

$n_{HCl} = \frac{PV}{RT} = \frac{1 \text{ atm} \times 448 \text{ L}}{0.0821 \text{ L atm/mol K} \times 546 \text{ K}} = \frac{448}{44.8326} \approx 10 \text{ mol}$

3. Determine the limiting reagent:

The balanced equation shows that 1 mole of $MnO_2$ reacts with 4 moles of HCl.

• For $MnO_2$: $\frac{\text{initial moles}}{\text{stoichiometric coefficient}} = \frac{3}{1} = 3$
• For HCl: $\frac{\text{initial moles}}{\text{stoichiometric coefficient}} = \frac{10}{4} = 2.5$

Since 2.5 < 3, HCl is the limiting reagent.

Evaluate the options:

(A) $MnO_2$ is limiting reagent.

This statement is incorrect. HCl is the limiting reagent.

(B) Chlorine gas produced contains $15 \times 10^{23}$ molecules.

Based on the limiting reagent (HCl):

From the stoichiometry, 4 moles of HCl produce 1 mole of $Cl_2$.

Moles of $Cl_2$ produced = $10 \text{ mol HCl} \times \frac{1 \text{ mol } Cl_2}{4 \text{ mol HCl}} = 2.5 \text{ mol } Cl_2$.

Number of $Cl_2$ molecules = moles $\times N_A = 2.5 \text{ mol} \times 6 \times 10^{23} \text{ molecules/mol} = 15 \times 10^{23}$ molecules.

This statement is correct.

(C) Moles of excess reactant left is 0.5 moles.

The excess reactant is $MnO_2$.

Moles of $MnO_2$ reacted = $10 \text{ mol HCl} \times \frac{1 \text{ mol } MnO_2}{4 \text{ mol HCl}} = 2.5 \text{ mol } MnO_2$.

Moles of excess $MnO_2$ left = Initial moles - Reacted moles = $3 \text{ mol} - 2.5 \text{ mol} = 0.5 \text{ mol}$.

This statement is correct.

(D) If % yield of reaction is 50%, then mass of $MnCl_2$ obtained will be 315.

First, calculate the theoretical mass of $MnCl_2$ produced:

From stoichiometry, 4 moles of HCl produce 1 mole of $MnCl_2$.

Moles of $MnCl_2$ produced = $10 \text{ mol HCl} \times \frac{1 \text{ mol } MnCl_2}{4 \text{ mol HCl}} = 2.5 \text{ mol } MnCl_2$.

Theoretical mass of $MnCl_2 = 2.5 \text{ mol} \times 126 \text{ g/mol} = 315 \text{ gm}$.

If the % yield is 50%, the actual mass of $MnCl_2$ obtained would be:

Actual mass = Theoretical mass $\times \frac{\% \text{ yield}}{100} = 315 \text{ gm} \times \frac{50}{100} = 157.5 \text{ gm}$.

This statement is incorrect.

Conclusion:

Statements (B) and (C) are correct.

Explanation of the solution:

1. Calculated initial moles of $MnO_2$ from its mass and molar mass.
2. Calculated initial moles of HCl gas using the Ideal Gas Law (PV=nRT) as conditions were not STP.
3. Determined the limiting reagent by comparing the ratio of initial moles to stoichiometric coefficients. HCl was found to be the limiting reagent.
4. Used the moles of the limiting reagent (HCl) to calculate the theoretical moles and number of molecules of $Cl_2$ produced, verifying option (B).
5. Used the moles of the limiting reagent (HCl) to calculate the moles of $MnO_2$ consumed and subsequently the moles of excess $MnO_2$ left, verifying option (C).
6. Calculated the theoretical mass of $MnCl_2$ produced. Then, calculated the actual mass of $MnCl_2$ at 50% yield, disproving option (D).

Answer: B, C