Question

# What mass of MnO2 is required to produce 1.12 litres of chlorine gas at STP acccording to the given equation? MnO2 (s)+ 4HCl (aq)→MnCl2 (aq)+ 2H2O (l)+ Cl2 (g) Atomic masses of H = 1, Cl = 35.5, Mn = 55, O = 16

A
4.35 g
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B
25.2 g
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C
6.1 g
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D
25.9 g
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Solution

## The correct option is A 4.35 gThe molecular mass ofMnO2 = 55 + 32 = 87 g Also, 1 mole of gas at STP acquires 22.4 litre volume. The given equation is: MnO2 (s)+ 4HCl (aq)→MnCl2 (aq)+ 2H2O (l)+ Cl2 (g) From the above equation, 1 mole of MnO2 is required to produce 1 mole of Cl2. ∴ 87 g of MnO2 is required to produce 22.4 L of Cl2 at STP ∴ For 1.12 L of Cl2 = 1.12×8722.4 = 4.35 g ofMnO2 is required.

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