Determine the volume of 0.175 M sulfuric acid required for interaction with barium chloride contained in 200 ml of 0.35 M solution

Determine the volume of 0.175 M sulfuric acid required for interaction with barium chloride contained in 200 ml of 0.35 M solution. Calculate the mass of the sediment formed.

H2SO4 + BaCl2 → BaSO4 ↓ + 2HCl
V (BaCl2) = 200 ml = 0.2 l
c (BaCl2) = 0.35M
0.35 = n / 0.2 ⇒ n = 0.35 x 0.2 = 0.07 mol
n (H2SO4) = n (BaCl2) = 0.07 mol
c (H2SO4) = 0.175M
V = n / c = 0.07 / 0.175 = 0.4 l
m (BaSO4) = 0.07 x (137 + 96) = 16.31 g
Answer: the volume of sulfuric acid is 0.4 l, the mass of the resulting precipitate is 16.31 g.

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