Question

Find mas of caco3 required to react with 40 ml, n/2 hcl

Answer 1

Answer:

1 mole of CaCO3 reacts with 2 moles of HCl from the equation.

We have 0.5M 4mL HCl i.e. no of moles of HCl is : 40/500 = 0.125

Thus no of moles required to completely consume 0.125 moles of HCl = 0.125/2 = 0.0625

Mass of CaCO3 = no of moles * molar mass = 0.0625*100 = 6.25g

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