How much precipitate was obtained by reacting 800 mg of a 30 percent NaOH solution with an excess of copper sulfate solution.

univerkov | November 29, 2020 | education

2NaOH + CuSO4 = Cu (OH) 2 + Na2SO4
m (NaOH) = 0.8 * 0.3 = 0.24 gr.
n (NaOH) = 0.24 / 40 = 0.006 mol
Copper hydroxide was formed 2 times less in moles, which means
n (Cu (OH) 2) = 0.003 mol
m (Cu (OH) 2) = 0.003 * 98 = 0.294 grams = 294 mg (in theory)
196/294 * 100% = 66.67%
Answer: 66.67%

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