When 8.74 g MnO₂ is treated with HCl, then what will be the weight of Cl₂(g) obtained? Molar mass of MnO₂ = 87.4 g/mol.

Correct Option/Answer:
The correct answer is 7.1 g.

Step-by-step Explanation:

Step 1: Write down the balanced chemical equation
When manganese dioxide (${\mathrm{MnO}}_2$) reacts with hydrochloric acid ($\mathrm{HCl}$), it forms manganese(II) chloride (${\mathrm{MnCl}}_2$), water ($H_{2}O$), and chlorine gas (${\mathrm{Cl}}_2$).

${\mathrm{MnO}}_2+4\mathrm{HCl}\to {\mathrm{MnCl}}_2+2H_{2}O+{\mathrm{Cl}}_2$

Step 2: Determine the molar mass of each component
Given molar mass of ${\mathrm{MnO}}_2=87.4 \mathrm{g/mol}$
Molar mass of chlorine gas (${\mathrm{Cl}}_2$):

$\text{Molar mass of}{\mathrm{Cl}}_2=2\times 35.5=71.0 \mathrm{g/mol}$

Step 3: Calculate the number of moles of ${\mathrm{MnO}}_2$ given
The formula for moles is:

$\text{Moles of}{\mathrm{MnO}}_2=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Moles of}{\mathrm{MnO}}_2=\frac{8.74 \text{g}}{87.4 \text{g/mol}}=0.1 \text{mol}$

Step 4: Use stoichiometry to find the weight of ${\mathrm{Cl}}_2$ formed
From the balanced equation, 1 mole of ${\mathrm{MnO}}_2$ produces 1 mole of ${\mathrm{Cl}}_2$ gas.
Therefore, 0.1 mol of ${\mathrm{MnO}}_2$ will produce 0.1 mol of ${\mathrm{Cl}}_2$ gas.

Now, calculate the mass of 0.1 mol of ${\mathrm{Cl}}_2$:

$\text{Mass of}{\mathrm{Cl}}_2=\text{Moles}\times \text{Molar mass}$

$\text{Mass of}{\mathrm{Cl}}_2=0.1 \text{mol}\times 71.0 \text{g/mol}=7.1 \text{g}$

Thus, the weight of ${\mathrm{Cl}}_2$ gas obtained is 7.1 g.