Question

What mass of ${\text{MnO}}_2$ is required to produce 1.12 litres of chlorine gas at STP acccording to the given equation?

${\text{MnO}}_2\text{(s)}+\text{4HCl}\text{(aq)}\to {\text{MnCl}}_2\text{(aq)}+{\text{2H}}_2\text{O}\text{(l)}+{\text{Cl}}_2\text{(g)}$

Atomic masses of H = 1, Cl = 35.5, Mn = 55, O = 16

• A. 4.35 g
• B. 25.2 g
• C. 6.1 g
• D. 25.9 g

Answer 1

The correct option is A 4.35 g

The molecular mass of${\text{MnO}}_2$ = 55 + 32 = 87 g
Also, 1 mole of gas at STP acquires 22.4 litre volume.
The given equation is:

${\text{MnO}}_2\text{(s)}+\text{4HCl}\text{(aq)}\to {\text{MnCl}}_2\text{(aq)}+{\text{2H}}_2\text{O}\text{(l)}+{\text{Cl}}_2\text{(g)}$

From the above equation, 1 mole of ${\text{MnO}}_2$ is required to produce 1 mole of ${\text{Cl}}_2$.

$\therefore$ 87 g of ${\text{MnO}}_2$ is required to produce 22.4 L of ${\text{Cl}}_2$ at STP

$\therefore$ For 1.12 L of ${\text{Cl}}_2$ = $\frac{1.12\times 87}{22.4}$ = 4.35 g of${\text{MnO}}_2$ is required.