дано
n(aL) = 3 mol
V(H2)-?
2Al+6HCL-->2AlCL3+3H2
2n(Al) = 3(H2)
n(H2) = 3*3/2 = 4.5mol
V(H2) = n*Vm = 4.5 * 22.4 = 100.8 L
ответ 100.8 л
