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.