What is the mass of salt obtained by the interaction of 50 g of calcium with 98 g of sulfuric acid?

univerkov | February 28, 2021 | education

Data: m (Ca) = 50 g; m (H2SO4) = 98 g.

1) The equation of the considered reaction: Ca (calcium) + H2SO4 (acid) = CaSO4 (obtained salt) + H2 ↑ (hydrogen).

2) Molar masses of substances: M (Ca) = 40 g / mol; M (H2SO4) = 2 * 1 (hydrogen) + 32 (sulfur) + 4 * 16 (oxygen) = 98 g / mol; M (CaSO4) = 40 (calcium) + 32 (sulfur) + 4 * 16 (oxygen) = 136 g / mol.

3) Calculated amounts of substance:

ν (Ca) = 50 g / 40 g / mol = 1.25 mol.

ν (H2SO4) = 98 g / 98 g / mol = 1 mol.

Lack of sulfuric acid.

4) Proportion: m (H2SO4) / M (H2SO4) = ν (1 mol) = m (CaSO4) / M (CaSO4).

m (CaSO4) = 98 * 136/98 = 136 g.

Answer: You should have got 136 g of salt.

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