Hydrochloric acid was applied to a solution weighing 200 g with a mass fraction of sodium silicate
Hydrochloric acid was applied to a solution weighing 200 g with a mass fraction of sodium silicate of 6.1%. Calculate the mass of the precipitate formed.
1. Let’s write down the equation of the proceeding reaction:
Na2SiO3 + 2HCl = H2SiO3 ↓ + 2NaCl;
2. find the mass of sodium silicate:
m (Na2SiO3) = w (Na2SiO3) * m (solution) = 0.061 * 200 = 12.2 g;
3.Calculate the chemical amount of silicate:
n (Na2SiO3) = m (Na2SiO3): M (Na2SiO3);
M (Na2SiO3) = 23 * 2 + 28 + 3 * 16 = 122 g / mol;
n (Na2SiO3) = 12.2: 122 = 0.1 mol;
4.determine the amount of silicic acid precipitate:
n (H2SiO3) = n (Na2SiO3) = 0.1 mol;
5.Calculate the mass of the sediment:
m (H2SiO3) = n (H2SiO3) * M (H2SiO3);
M (H2SiO3) = 2 + 28 + 3 * 16 = 78 g / mol;
m (H2SiO3) = 0.1 * 78 = 7.8 g.
Answer: 7.8 g.