Question 55

Aqueous HCI reacts with $$MnO_{2} \left(s\right)$$ to form $$MnCl_{2}\left(aq\right)$$, $$Cl_{2}\left(g\right)$$ and $$H_{2}O\left(l\right)$$. What is the weight (in g) of Cl_{2} liberated when 8.7 g of $$MnO_{2} \left(s\right)$$ is reacted with excess aqueous HCI solution ?
(Given Molar mass in g $$mol^{-1}$$ Mn = 55, Cl = 35.5, 0 = 16, H = l )

Reaction: $$MnO_2 + 4HCl \to MnCl_2 + Cl_2 + 2H_2O$$

Molar mass of $$MnO_2 = 55 + 32 = 87$$ g/mol.

Moles of $$MnO_2 = \frac{8.7}{87} = 0.1$$ mol.

From stoichiometry: 1 mol $$MnO_2$$ → 1 mol $$Cl_2$$.

Moles of $$Cl_2 = 0.1$$ mol.

Mass of $$Cl_2 = 0.1 \times 71 = 7.1$$ g.

The answer is Option 4: 7.1 g.

Create a FREE account and get:

Download resources