Дано
m(AL) = 27 g
m(ppaHCL) = 50 g
W(HCL) = 15%
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V(H2)-?
M(Al)= 27 g/mol
n(AL) = m/M= 27/27 = 1 mol
m(HCL) = 50*15% / 100% = 7.5 g
M(HCL) = 36.5 g/mol
n(HCL) = m/M= 7.5 / 36.5 = 0.2 mol
n(Al) > n(HCL)
0.2mol X
6AL+6HCL-->2AlCL3+3H2
6 mol 3*22.4
X = 0.2*3*22.4 / 6 = 2.24 L
ответ 2.24 л