m (AlCl3) = 24 g
m (NAOH) = 80 g
m (draft) -?
AlCl3 + 3NaOH = 3NaCl + Al (OH) 3, – we solve the problem using the composed reaction equation:
1) Find the amount of aluminum chloride and sodium hydroxide:
n (AlCl3) = m: M = 24 g: 133.5 g / mol = 0.18 mol
n (NaOH) = m: M = 80 g: 40 g / mol = 2 mol
We start from a lower value to get more accurate calculations. We work with AlCl3:
2) We compose a logical expression:
if 1 mol of AlCl3 gives 1 mol of Al (OH) 3,
then 0.18 mol of AlCl3 will give x mol of Al (OH) 3,
then x = 0.18 mol.
3) Find the mass of aluminum hydroxide precipitated during the reaction:
m (Al (OH) 3) = n * M = 0.18 mol * 78 g / mol = 14.04 g.
Answer: m (Al (OH) 3) = 14.04 g.
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