# The mass of aluminum hydroxide formed during the interaction of a solution

The mass of aluminum hydroxide formed during the interaction of a solution containing 24 grams of sodium hydroxide and 26.7 grams of aluminum chloride is equal.

1. The reaction equation has the form:

3NaOH + AlCl3 = Al (OH) 3 + 3NaCl;

2. find the chemical amount of sodium hydroxide:

n (NaOH) = m (NaOH): M (NaOH);

M (NaOH) = 23 + 17 = 40 g / mol;

n (NaOH) = 24:40 = 0.6 mol;

3.calculate the amount of aluminum chloride

n (AlCl3) = m (AlCl3): M (AlCl3);

M (AlCl3) = 27 + 71 = 133.5 g / mol;

n (AlCl3) = 26.7: 133.5 = 0.2 mol;

4.set the amount of aluminum hydroxide

n (Al (OH) 3) = n (NaOH): 3 = 0.6: 3 = 0.2 mol;

5.determine the mass of the obtained hydroxide

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

M (Al (OH) 3) = 27 + 51 = 78 g / mol;

m (Al (OH) 3) = 0.2 * 78 = 15.6 g.

Answer: 15.6 g.

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