An excess of calcium nitrate solution was added to a potassium silicate solution weighing 25.5 g and a mass fraction of 10%.

An excess of calcium nitrate solution was added to a potassium silicate solution weighing 25.5 g and a mass fraction of 10%. Calculate the mass of the precipitate formed.

1) Let’s compose the reaction equation:
K2SiO3 + Ca (NO3) 2 = CaSiO3 ↓ + 2KNO3
2) Calculate the mass and amount of the potassium silicate substance contained in the solution:
m (K2SiO3) = (m (solution) ∙ ω) / 100% = (25.5 ∙ 10) / 100 = 2.55 g;
n (K2SiO3) = m (K2SiO3) / M (K2SiO3) = 2.55 / 154 = 0.0166 mol.
3) Determine the mass of the sediment:
according to the reaction equation
n (K2SiO3) = n (CaSiO3) = 0.0166 mol;
m (CaSiO3) = n (CaSiO3) ∙ M (CaSiO3) = 0.0166 ∙ 116 = 1.93 g.
Answer: 1.93 g.

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