The reaction of MnO2 and HCl produces 2.24 L of Cl2 gas. What mass of MnO2 was used up by the reaction? (At. Mass of Mn = 55gmol)
MnO2+4HCl→2H2O+MnCl2+Cl2

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7.19g

8.7g

2.24g

18.69g

answer is B.

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Detailed Solution

When one mole of manganese dioxide (MnO2) reacts with 4 moles of hydrogen chloride (HCl), two moles of water (H2O), one mole of manganese chloride (MnCl2) and one mole of Chlorine gas is formed.
The following reaction is shown by the given equation:
MnO2+4HCl→2H2O+MnCl2+Cl2
At standard temperature and pressure i.e., at S.T.P 1 mole of any substance produces a volume of 22.4 L. Thus, 1 mole of MnO2 produces a volume of 22.4 L of Cl2 as 1 mole of Cl2 is formed from 1 mole of MnO2. Now, we’ll calculate molar mass of MnO2 where molar or atomic mass of Mn is given as 55 g mol and atomic mass of oxygen is 16 g mol.
Molar mass of MnO2 (M) = 1 × atomic mass of Mn + 2 × atomic mass of O.
M=1×55+2×16
⇒M=55+32
⇒M=87g
Thus, the amount of mass of MnO2 required to produce 22.4 L of Cl2 gas = 87g as 1 mole of MnO2 = 87g.
Amount of mass of MnO2 required to produce 1 L of Cl2 gas = 8722.4.
So, the amount of mass of MnO2 required to produce 2.24 L of Cl2 gas = 87×2.2422.4i.e.,8710.
Hence, the amount of mass of MnO2 used up by the reaction to produce 2.24 L of Cl2 gas is 8710 i.e., 8.7g.

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