In the reaction, 522.49 moles of barium chloride and unlimited amount of sulfuric acid react, to give, as products barium sulfate and hydrochloric acid. How much is the minimum amount of reactant required of sulfuric acid in moles. Then additionally, how much product is produced of barium sulfate, in grams, and hydrochloric acid in grams?

Question

Dominguez

Chemistry

28 May, 20:33

In the reaction, 522.49 moles of barium chloride and unlimited amount of sulfuric acid react, to give, as products barium sulfate and hydrochloric acid. How much is the minimum amount of reactant required of sulfuric acid in moles. Then additionally, how much product is produced of barium sulfate, in grams, and hydrochloric acid in grams?

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Answers (1)

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Kevin Bright · Answer 1

1) Answer is: the minimum amount of reactant required of sulfuric acid is 522.49 moles.

Balanced chemical reaction:

BaCl₂ (aq) + H₂SO₄ (aq) → BaSO₄ (s) + 2HCl (aq).

n (BaCl₂) = 522.49 mol; amount of barium chloride.

From chemical reaction: n (BaCl₂) : n (H₂SO₄) = 1 : 1.

n (H₂SO₄) = n (BaCl₂).

n (H₂SO₄) = 522.49 mol; amount of sulfuric acid.

2) Answer is: mass of barium sulfate is 121949.16 grams.

From chemical reaction: n (BaCl₂) : n (BaSO₄) = 1 : 1.

n (BaSO₄) = n (BaCl₂).

n (BaSO₄) = 522.49 mol; amount of barium sulfate.

m (BaSO₄) = n (BaSO₄) · M (BaSO₄).

m (BaSO₄) = 522.49 mol · 233.4 g/mol.

m (BaSO₄) = 121949.16 g; mass of barium sulfate.

3) Answer is: mass of hydrochloric acid is 38089.52 grams.

From chemical reaction: n (BaCl₂) : n (HCl) = 1 : 2.

n (HCl) = 2 · n (BaCl₂).

n (HCl) = 2 · 522.49 mol.

n (HCl) = 1044.98 mol; amount of hydrochloric acid.

m (HCl) = n (HCl) · M (HCl).

m (HCl) = 1044.98 mol · 36.45 g/mol.

m (HCl) = 38089.52 g; mass of hydrochloric acid.