How many grams of sodium hydroxide is required to convert 16 g copper (II) sulfate to copper (II) hydroxide?

Given:
m (CuSO4) = 16 g

To find:
m (NaOH) -?

Solution:
1) CuSO4 + 2NaOH => Cu (OH) 2 ↓ + Na2SO4;
2) M (CuSO4) = Mr (CuSO4) = Ar (Cu) * N (Cu) + Ar (S) * N (S) + Ar (O) * N (O) = 64 * 1 + 32 * 1 + 16 * 4 = 160 g / mol;
3) n (CuSO4) = m (CuSO4) / M (CuSO4) = 16/160 = 0.1 mol;
4) n (NaOH) = n (CuSO4) * 2 = 0.1 * 2 = 0.2 mol;
5) M (NaOH) = Mr (NaOH) = Ar (Na) * N (Na) + Ar (O) * N (O) + Ar (H) * N (H) = 23 * 1 + 16 * 1 + 1 * 1 = 40 g / mol;
6) m (NaOH) = n (NaOH) * M (NaOH) = 0.2 * 40 = 8 g.

Answer: The mass of NaOH is 8 g.



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