To 200 g of a 13.35% solution of aluminum chloride was added 300 g of a 12.6% solution of sodium sulfite.

| April 27, 2021 | education

To 200 g of a 13.35% solution of aluminum chloride was added 300 g of a 12.6% solution of sodium sulfite. Find the mass of the precipitate obtained and the mass of the solution after the reaction.

m1 solution = 200 g.

W1 solution = 13.35%.

m2 size = 300 g.

W2 solution = 12.6%.

Let us determine the mass of the resulting precipitate and the mass of the solution after the reaction.

We write down the equation of reactions.

2AlCl3 + 3Na2SO3 = Al2 (SO3) 3 + 6NaCl.

We find the mass of aluminum chloride and sodium sulfite. We write down the solution.

W = m r.v / m r-ra * 100%.

m (AlCl3) = 200 * 0.1335 = 26.7 g.

m (Na2SO3) = 300 * 0.126 = 37.8 g.

Determine what substance is in excess.

M (AlCl3) = 133.34 g / mol.

M (Na2SO3) = 126.03 g / mol.

26.7 g AlCl3 – x g

2 * 133.34 – 3 * 126.03

X = 26.7 * 3 * 126.3: (2 * 133.34) = 37.9 sodium sulfite.

This means that sodium sulfite is in short supply. We count by sodium sulfite.

37.8 g Na2sO3 – x Al2 (SO3) 3

3 * 126.03 g / mol Na2SO3 – 294 g / mol Al2 (SO3) 3

X = 37.8 * 294: (3 * 126.03) = 29.4 g.

This means that the mass of the precipitate formed is 29.4 grams.

m solution = 200 + 300 = 500 g.

Answer: m = 29.4 g.



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