дано

n(AlCL3) = 1.5 mol

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V(CL2)-?

2Al+3CL2-->2ALCL3

3n(CL2) = 2n(AlCL3)

n(CL2) = 3*1.5 / 2 = 2.25 mol

V(CL2) = n(CL2)*Vm = 2.25 * 22.4 = 50.4 L

ответ 50.4 л