What amount of 25% H2SO4 will be required to react
completely with 100 g (20% impure) CaCO3?

Dear Student,

We have been given the amount of CaCO3 present and the percentage of H2SO4 present in the solution.

Weight of CaCO3 = 100gIt is given that 20% CaCO3 is impure. So, amount of CaCO3 which is pure and can react is 80 g.Molar mass of CaCO3=100 g mol-1Number of moles of CaCO3 = 80100=0.8Now, the reaction is as follows:H2SO4 + CaCO3  CaSO4 + CO2 + H2O1 mole of CaCO3 reacts with 1 mole of H2SO4Thus, 0.8 mole of CaCO3 reacts with 0.8 mole of H2SO4(molar mass 98 g)0.8 mole contains 0.8 x 98 = 78.4 g of H2SO4Thus, amount of 25% H2SO4 required is : 100×78.425= 313.6 g

Thus, the amount required is 313.6 g.

Hope it helps.


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