A solution containing 3.48 g of aluminum sulfate was added to a solution containing 0.4 magnesium hydroxide.

| August 6, 2021 | education

A solution containing 3.48 g of aluminum sulfate was added to a solution containing 0.4 magnesium hydroxide. indicate the formula of the substance taken in deficiency and the mass of the resulting precipitate (g)

1. Al2 (SO4) 3 + 3Mg (OH) 2 = 2Al (OH) 3 + 3MgSO4;

2. find the chemical amount of magnesium sulfate:

n (Al2 (SO4) 3) = m (Al2 (SO4) 3): M (Al2 (SO4) 3);

M (Al2 (SO4) 3) = 2 * 27 + 3 * 32 + 4 * 3 * 16 = 342 g / mol;

n (Al2 (SO4) 3) = 3.48: 342 = 0.0102 mol;

3. since n (Mg (OH) 2) = 0.4 mol, then aluminum sulfate is in short supply, we determine the chemical amount of aluminum hydroxide:

n (Al (OH) 3) = n (Al2 (SO4) 3) * 2 = 0.0102 * 2 = 0.0204 mol;

4.calculate the mass of the sediment:

m (Al (OH) 3) = n (Al (OH) 3) * M (Al (OH) 3);

M (Al (OH) 3) = 27 + 3 * 17 = 78 g / mol;

m (Al (OH) 3) = 0.0204 * 78 = 1.59 g.

Answer: 1.59 g.



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