Silicon oxide 4 weighing 30 grams was fused with 30 grams of sodium hydroxide to determine the mass of sodium silicate that can be obtained.
Let’s execute the solution:
In accordance with the condition of the problem, we compose the equation of the process:
SiO2 + 2NaOH = Na2SiO3 + H2O – ion exchange, sodium silicate obtained;
M (SiO2) = 60 g / mol;
M (NaOH) = 39.9 g / mol;
M (Na2SiO3) = 121.8 g / mol.
Let’s determine the amount of starting materials:
Y (SiO2) = m / M = 30/60 = 0.5 mol (deficient substance);
Y (Na2SiO3) = 0.5 mol since the amount of substances is 1 mol;
Y (NaOH) = m / M = 30 / 39.9 = 0.75 mol (substance in excess).
Calculations are carried out for the substance in deficiency.
Find the mass of the product:
m (Na2SiO3) = Y * M = 0.5 * 121.8 = 60.9 g
Answer: the mass of sodium silicate is 60.9 g
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