дано
m(Al) = 5,4 g
V(CL2) = ?
2Al+3CL2-->2AlCL3
M(AL) = 27 g/mol
n(Al) = m/M = 5.4 / 27 = 0.2 mol
2n(Al) = 3n(CL2)
n(CL) = 0.2*3 / 2 = 0.3 mol
V(CL2) = n*Vm = 0.3 * 22,4 = 6.72 L
ответ 1) 6,72л
Объяснение:
