a)\ 2Al+6HCl->2AlCl_3+3H_2

b)

n_{H_2}={6,72}/{22,4}=0,3(mol)

Theo PT: n_{Al}=2/3n_{H_2}=0,2(mol)

->\%m_{Al}={0,2.27}/{15}.100\%=36\%

->\%m_{Cu}=100-36=64\%

c)

Theo PT: n_{AlCl_3}=n_{Al}=0,2(mol)

Có m_{dd\ spu}=0,2.27+592,2-0,3.2=597(g)

->C\%_{AlCl_3}={0,2.133,5}/{597}.100\%\approx 4,47\%

a,PTHH:2Al+6HCl->2AlCl_3+3H_2

b,n_{H_2}=(6,72)/(22,4)=0,3(mol)

Theo PTHH: n_{Al}=2/3n_{H_2}=0,2(mol)

=>\%m_{Al}=(0,2.27)/(15).100\%=36\%

=>\%m_{Cu}=100\%-36\%=64\%

c,

Theo PTHH: n_{AlCl_3}=n_{H_2}=0,2(mol)

m_{ddsau}=0,2.27+592,2-0,3.2=597(g)

=>C\%_{ddAlCl_3}={0,2.133,5}/{597}.100\%=4,472\%