Chlorine is prepared in the laboratory by treating manganese dioxide $({M n O}_{2})$ with aqueous hydrochloric acid according to the reaction:

$4 H C l (a q) + {M n O}_{2} (s) \to 2 H_{2} O (I) + {M n C l}_{2} (a q) + {C l}_{2} (g)$ .

How many grams of $H C l$ react with 5.0 g of manganese dioxide ?

Open in App

Solution

The balanced chemical equation is:

$M n O_{2} + 4 H C l \to M n C l_{2} + 2 H_{2} O + 5 C l_{2}$ .

The molar mass of $M n O_{2}$ is $55 + 2 \times 16 = 87$ g/mol.

The molar mass of $H C l$ is $1 + 35.5 = 36.5$ g/mol.

Hence, 5 g of $M n O_{2}$ will react with $\frac{36.5 \times 4 \times 5}{87} = 8.39$ g of HCl.

Suggest Corrections

Similar questions

(M n O_{2})

M n O_{2} + 4 H C l \to M n C l_{2} + C l_{2} + 2 H_{2} O

M n O_{2}

M O_{2})

M n O_{2} + 4 H C l \to M n C l_{2} + C l_{2} + 2 H_{2} O

5 g M n O_{2}

F e

M n O_{2}

Join BYJU'S Learning Program