Дано
m(AL)=5 g
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V(H2)-?
m(ALCL3)-?
5 X
6HCL+2Al-->2ALCL3+3H2 Vm=22.4 L/mol
2*27 3*22.4
X=5*67.2/54=6.2 L
M(Al)=27 g/mol
n(Al)=m/M=5/27=0.185 mol
n(Al)=n(AlCL3)=0.185 mol
M(ALCL3)=133,5 g/mol
m(AlCL3)=n*M=0.185*133.5= 24.7 g
ответ 6.2 л , 24.7 г