Дано
m(AL)=2.7 g
m(ppaHCL)=50 g
W(HCL)=15%
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V(H2)-?
m(HCL)=m(ppa)*W(HCL)/100% = 50*15%/100% = 7.5 g
2.7 7.5 X
2Al+6HCL-->2AlCL3+3H2 Vm=22.4 L/mol
2*27 3*22.4
M(Al)=27 g/mol
n(Al)=m/M =2.7/27 = 0.1 mol
M(HCL)=36.5 g/mol
n(HCL)=m/M=7.5/36.5 = 0.2 mol
n(AL)2.7/54 = X / 67.2
X=3.36 L
ответ 3.36 л