# Calculate the mass of the precipitate that forms upon the interaction of 450 g of a 15% tin sulfate

Calculate the mass of the precipitate that forms upon the interaction of 450 g of a 15% tin sulfate solution with a potassium hydroxide solution

Data: mр is the mass of the taken solution (mр = 450 g); ωSnSO4 – tin sulfate content (ωSnSO4 = 15% = 0.15).

Const: MSnSO4 – molar mass of tin sulfate (MSnSO4 = 214.77 g / mol); MSn (OH) 2 – molar mass of tin hydroxide (MSn (OH) 2 = 152.72 g / mol).

1) The mass of the reacted tin sulfate: mSnSO4 = ωSnSO4 * mр = 0.15 * 450 = 67.5 g.

2) Amount of tin sulfate substance: νSnSO4 = mSnSO4 / MSnSO4 = 67.5 / 214.77 = 0.3143 mol.

3) Reaction equation: Sn (OH) 2 (tin hydroxide) + K2SO4 (potassium sulfate) = SnSO4 (potassium sulfate) + 2KOH (potassium hydroxide).

4) Amount of tin hydroxide substance: νSn (OH) 2 = νSnSO4 = 0.3143 mol.

5) The mass of the resulting gel-like white precipitate (tin hydroxide): mSn (OH) 2 = νSn (OH) 2 * MSn (OH) 2 = 0.3143 * 152.72 ≈ 48 g.

Answer: The mass of the resulting gel-like white precipitate is 48 g.

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