Question

What is the volume of 0.1N hcl required To react completely with 1.0g of pure calcium carbonate

Answer 1

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Vol = w x 1000/ 50 x 0.1

The value 50 is there because eq. weight of CaCO₃ = 100/2 (2 electrons are taken away from Ca)

Then, 1 x 1000/50 x 0.1 = 1000/5 = 200 ml

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