What is the mass of the precipitate formed: when an excess of barium chloride solution interacts with 49 g of sulfuric acid?
| July 13, 2021 | education
Given:
m (H2SO4) = 49 g
To find:
m (draft) -?
Decision:
1) BaCl2 + H2SO4 => 2HCl + BaSO4 ↓;
2) n (H2SO4) = m (H2SO4) / Mr (H2SO4) = 49/98 = 0.5 mol;
3) n (BaSO4) = n (H2SO4) = 0.5 mol;
4) m (BaSO4) = n (BaSO4) * Mr (BaSO4) = 0.5 * 233 = 116.5 g.
Answer: The mass of BaSO4 is 116.5 g.
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