# (solved)Question 1.36 of NCERT Class XI Chemistry Chapter 1

Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction 4 HCl (aq) + MnO2 (s) → 2H2O (l) + MnCl2 (aq) + Cl2 (g) How many grams of HCl react with 5.0 g of manganese dioxide?
(Rev. 19-Dec-2022)

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### Question 1.36 NCERT Class XI Chemistry

Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction 4 HCl (aq) + MnO2 (s) → 2H2O (l) + MnCl2 (aq) + Cl2 (g) How many grams of HCl react with 5.0 g of manganese dioxide?

### Video Explanation(detailed solution given after this video)

Please watch this youtube video for a quick explanation of the solution:

### Solution in Detail

$\displaystyle \underline{\underline{\text{Step 1: Get mols of MnO2}}}$

Atomic mass of Mn = 55u, O = 16u

$\displaystyle \therefore$ molar mass M = 55 + (16 x 2) = 87 g/mol

we know $\displaystyle n = \frac{\text{w}}{\text{M}}$

Given $\displaystyle w = 5\text{ g MnO}_2$

$\displaystyle \therefore n = \bigg(\frac{5}{87}\bigg) \text{ mol MnO}_2$

$\displaystyle \underline{\underline{\text{Step 2: Get mols of HCl}}}$

The given equation gives

1 mol MnO2 reacts with 4 mol HCl

$\displaystyle \therefore \frac{5}{87}$ reacts with $\displaystyle \frac{5}{87} \times 4 = \bigg(\frac{20}{87}\bigg)$ mol HCl

$\displaystyle \underline{\underline{\text{Step 3: Get grams of HCl}}}$

Molar mass of HCl = 1 + 35.5 = 36.5 (g/mol)

but $\displaystyle n = \frac{w}{M}$

$\displaystyle \therefore w = n \times M$

$\displaystyle \therefore w = \bigg(\frac{20}{87}\bigg) \times 36.5 = 8.40\text{ g}\:\underline{Ans}$