m(Cu+Al)=6грV(H2)=3.7лw-?2Al+6HCl=2AlCl3+3H2n(H2)=3.7/22.4=0.165моль2к3 или 1к1.5 значитn(Al)=0.165/1.5=0.11мольm(al)=0.11*27=2.97 грm(Cu)=6-2.97=3.03грw(Al)=2.97/6*100=49.5%w(Cu)=3.03/6*100=50.5%
m(Cu+Al)=6гр
V(H2)=3.7л
w-?
2Al+6HCl=2AlCl3+3H2
n(H2)=3.7/22.4=0.165моль
2к3 или 1к1.5 значит
n(Al)=0.165/1.5=0.11моль
m(al)=0.11*27=2.97 гр
m(Cu)=6-2.97=3.03гр
w(Al)=2.97/6*100=49.5%
w(Cu)=3.03/6*100=50.5%