Calculate the mass of the precipitate, which is formed by the interaction of 160 g of a 15% solution of copper sulfate 2 with an excess of sodium hydroxide.
m (CuSO4) = 160 g
w% (CuSO4) = 15%
m (draft) -?
CuSO4 + 2NaOH = Cu (OH) 2 + Na2SO4, – we solve the problem based on the composed reaction equation:
1) Find the mass of copper sulfate that has reacted:
m (CuSO4) = 160 g * 0.15 = 24 g
2) Find the amount of copper sulfate:
n (CuSO4) = m: M = 24 g: 160 g / mol = 0.15 mol
3) We compose a logical expression:
If 1 mol of CuSO4 gives 1 mol of Cu (OH) 2,
then 0.15 mol of CuSO4 will give x mol of Cu (OH) 2,
then x = 0.15 mol.
4) Find the mass of precipitated copper hydroxide:
m (Cu (OH) 2) = n * M = 0.15 mol * 98 g / mol = 14.7 mol
Answer: m (Cu (OH) 2) = 14.7 mol.
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