A solution containing 13.6 g of zinc chloride was treated with a solution containing 10 g

A solution containing 13.6 g of zinc chloride was treated with a solution containing 10 g of sodium hydroxide. What is the mass of the precipitate formed as a result of the interaction?

Let’s implement the solution:

According to the condition of the problem, we write down the equation of the process:
ZnCl2 + 2NaOH = Zn (OH) 2 + 2NaCl – ion exchange, zinc hydroxide precipitate is formed;

We make calculations:
M (ZnCl2) = 136.3 g / mol;

M (NaOH) = 39.9 g / mol;

MZn (OH) 2 = 99.3 g / mol.

Let’s determine the amount of starting substances:
Y (ZnCl2) = m / M = 13.6 / 136.3 = 0.09 mol (deficient substance);

Y (NaOH) = m / M = 10 / 39.9 = 0.25 mol (substance in excess).

Calculations are made for the substance in deficiency.

We find the mass of the sediment:
M Zn (OH) 2 = Y * M = 0.09 * 99.3 = 8.94 g

Answer: the mass of zinc hydroxide is 8.94 g