# 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. Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.