45.7 grams of calcium chloride reacts with an excess of aluminum oxide. How many grams of aluminum chloride will be produced

1 answer:

5 0

Molar mass (CaCl2) = 40.1 +2*35.5 = 111.1 g/mol
Molar mass (AlCl3) = 27.0 +3*35.5= 133.5 g/ mol

3CaCl2+Al2O3 -------->3CaO +2AlCl3
mole from reaction 3 mol 2 mol
mass from reaction 3mol* 111.1g/mol 2 mol*133.5g/mol
333.3 g 267.0 g
mass from problem 45.7 g x g

Proportion:
333.3 g CaCl2 ------- 267.0 g AlCl3
45.7 g CaCl2 -------- x g AlCl3

x=45.7*267.0/333.3= 36.6 g AlCl3

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