Verified answer

дано

V(H2) = 11.2 L

---------------------

m(Al) - ?

2Al+6HCL-->2AlCL3+3H2

n(H2) = V / Vm = 11.2 / 22.4 = 0.5 mol

2n(AL)= 3n(H2)

n(AL) = 0.5 * 3 / 2 = 0.75 mol

M(Al) = 27 g/mol

m(Al) = n*M =0.75 * 27 = 20.25 g

ответ 20.25 г