Find the mass of the precipitate formed during the interaction of chromium (II) chloride and sodium hydroxide

Find the mass of the precipitate formed during the interaction of chromium (II) chloride and sodium hydroxide solution weighing 500 g with a mass fraction of the substance of 10%.

1.Let’s find the mass of sodium hydroxide in the solution.

500 g – 100%,

X g – 10%,

X = (500 g × 10%): 100%,

X = 50 g.

Or can be found by the formula:

w = m (substance): m (solution) x 100%.

hence m (substance) = (m (solution) × w): 100%.

m (substance) = (500 g × 10%): 100% = 50 g.

2.Let’s find the amount of sodium hydroxide substance.

n = m: M.

M (NaOH) = 23 + 16 + 1 = 40g / mol.

n = 500: 40 g / mol = 1.25 mol.

3. We compose the reaction equation, we find the quantitative ratios of substances.

CrCl3 + 2NaOH = Cr (OH) 2 + 2NaCl.

For 2 mol of sodium hydroxide there is 1 mol of chromium hydroxide, that is, the quantitative ratio is 2: 1. The amount of Cr (OH) 2 will be 2 times less than the amount of NaOH.

n (Cr (OH) 2) = 1.25 mol: 2 = 0.625 mol.

4. The mass of a substance is found by the formula:

m = n × M, where M is molar mass.

M ((Cr (OH) 2) = 86 g / mol.

m = 86 g / mol × 0.625 mol = 53.75 g.

Answer: 53.75 g.