Calculate the mass of aluminum hydroxide obtained by reacting 66.75 g of aluminum bromide and 33.6 g of potassium hydroxide.
m (AlBr3) = 66.75 g
m (KOH) = 33.6 g
m (Al (OH) 3) -?
1) AlBr3 + 3KOH => Al (OH) 3 + 3KBr;
2) n (AlBr3) = m / M = 66.75 / 267 = 0.25 mol;
3) n (KOH) = m / M = 33.6 / 56 = 0.6 mol;
4) n (Al (OH) 3) = n (KOH) / 3 = 0.6 / 3 = 0.2 mol;
5) m (Al (OH) 3) = n * M = 0.2 * 78 = 15.6 g.
Answer: The mass of Al (OH) 3 is 15.6 g.
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