Determine the mass of aluminum hydroxide formed by the interaction of 2 moles of aluminum chloride with 4 moles of sodium hydroxide
The problem for the excess and lack of reagents.
AlCl3 + 3NaOH = Al (OH) 3 + 3NaCl
According to the stoichiometry of the reaction, the substances react in a ratio of 1: 3, aluminum chloride in excess.
n (Al (OH) 3) = 1/3 n (NaOH) = 1.333 mol
m (Al (OH) 3) = n (Al (OH) 3) * M (Al (OH) 3) = 1.333 * 78 = 104 g.
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