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.
Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.