Calculate the mass of aluminum hydroxide that is formed by the interaction of aluminum sulfate with slaked lime weighing 7.4 g.
The equation for the reaction of calcium hydroxide (slaked lime) with aluminum sulfate:
3Ca (OH) 2 + Al2 (SO4) 3 = 2Al (OH) 3 + 3CaSO4
Amount of calcium hydroxide substance:
v (Ca (OH) 2) = m (Ca (OH) 2) / M (Ca (OH) 2) = 7.4 / 74 = 0.1 (mol).
According to the reaction equation, for 3 mol of Ca (OH) 2 2 mol of Al (OH) 3 are formed, therefore:
v (Al (OH) 3) = v (Ca (OH) 2) * 2/3 = 0.1 * 2/3 = 2/30 = 1/15 (mol).
Thus, the mass of the formed aluminum hydroxide is:
m (Al (OH) 3) = v (Al (OH) 3) * M (Al (OH) 3) = (1/15) * 78 = 5.2 (g).
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