Question

What mass of ${\text{MnO}}_2$ (in grams ) is required to produce 1.12 litres of chlorine gas at STP acccording to the given equation? (upto two decimals)

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

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

• A. 4.35

Answer 1

The correct option is A 4.35
The gram molecular mass of ${\text{MnO}}_2$ = 55 + 32 = 87 g

Also, 1 mole of gas at STP has a volume of 22.4 litre.

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 ${\text{Cl}}_2$.

i.e., 87 g of ${\text{MnO}}_2$ is required to produce 22.4 L of ${\text{Cl}}_2$ at STP

$\therefore$ Mass of ${\text{MnO}}_2$ required to produce 1.12 L of ${\text{Cl}}_2$ = $\frac{1.12\times 87}{22.4}$ = 4.35 g