The mass of 50%(w/w) solution of HCl required to react with 100 g of CaCO3 would be?

Engineering

Chemistry

Solution Stoichiometry

Question

The mass of 50%(w/w) solution of HCl required to react with 100 g of CaCO₃ would be?

73 g

100 g

146 g

200 g

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The reaction is:

2HCl + CaCO₃ → CaCl₂ + CO + 3H₂

100 g of CaCO₃ = 1 mole

So, Molar mass of HCl = 35.5 + 1 = 36.5

∴ 1 mole of HCl = 36.5 g

Since, we get 2 moles of HCl from 100 g(1 mole) of CaCO₃

Weight of HCl = 2(36.5) g = 73 g