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
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