Calculate the mass of the precipitate obtained by the interaction of a salt solution in which 16
February 6, 2021 | education
| Calculate the mass of the precipitate obtained by the interaction of a salt solution in which 16 g of copper (II) sulfate, with an excess of sodium hydroxide solution.
Let’s write the reaction equation:
CuSO4 + 2NaOH = Cu (OH) 2 ↓ + Na2SO4
Let’s find the amount of copper (II) sulfate substance:
v (CuSO4) = m (CuSO4) / M (CuSO4) = 16/160 = 0.1 (mol).
According to the reaction equation, from 1 mol of CuSO4, 1 mol of Cu (OH) 2 is formed, therefore:
v (Cu (OH) 2) = v (CuSO4) = 0.1 (mol).
Thus, the mass of the formed copper (II) hydroxide precipitate is:
m (Cu (OH) 2) = v (Cu (OH) 2) * M (Cu (OH) 2) = 0.1 * 98 = 9.8 (g).
Answer: 9.8 g.
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