Find the mass of the precipitate that forms when 320 g of a 20% copper sulfate solution

| September 8, 2021 | education

Find the mass of the precipitate that forms when 320 g of a 20% copper sulfate solution (2) interacts with 56 g of a 10% potassium hydroxide solution.

1. Let’s write down the reaction equation:

CuSO4 + 2KOH = Cu (OH) 2 + K2SO4.

2. Find the amount of copper sulfate and potassium hydroxide:

m (CuSO4) = m (solution) * ω (CuSO4) / 100% = 320 g * 20% / 100% = 64 g.

n (CuSO4) = m (CuSO4) / M (CuSO4) = 64 g / 160 g / mol = 0.4 mol.

m (KOH) = m (solution) * ω (KOH) / 100% = 56 g * 10% / 100% = 5.6 g.

n (KOH) = m (KOH) / M (KOH) = 5.6 g / 56 g / mol = 0.1 mol.

3. According to the reaction equation, it can be seen that potassium hydroxide is in short supply, therefore, from it we find the amount of precipitate:

n (Cu (OH) 2) = n (KOH) / 2 = 0.1 mol / 2 = 0.05 mol.

m (Cu (OH) 2) = n (Cu (OH) 2) * M (Cu (OH) 2) = 0.05 mol * 98 g / mol = 4.9 g.

Answer: m (Cu (OH) 2) = 4.9 g.



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