Câu20:Câu20:
TN1:TN1:
Đặt:nCu=x(mol),nZn=y(mol)Đặt:nCu=x(mol),nZn=y(mol)
mhh=135x+136y=81.36(g)(.)mhh=135x+136y=81.36(g)(.)
TN2:TN2:
Đặt:nCu=kx(mol),nZn=ky(mol),nAl=kz(mol)Đặt:nCu=kx(mol),nZn=ky(mol),nAl=kz(mol)
nX=kx+ky=0.5(mol)(1)nX=kx+ky=0.5(mol)(1)
nZnnAl=32⇒kykz=32⇒2ky−3kz=0(2)nZnnAl=32⇒kykz=32⇒2ky−3kz=0(2)
nH2=ky+1.5kz=13.4422.4=0.6(mol)(3)nH2=ky+1.5kz=13.4422.4=0.6(mol)(3)
(1),(2),(3):(1),(2),(3):kx=0.2,ky=0.3,kz=0.2kx=0.2,ky=0.3,kz=0.2
Từ(.):Từ(.):
⇒135(x+y)+y=81.36⇒135(x+y)+y=81.36
⇔135⋅0.5k+0.3k=81.36⇔135⋅0.5k+0.3k=81.36
⇔k=135⋅0.5+0.381.36=56⇔k=135⋅0.5+0.381.36=56
KĐ:KĐ:
x=0.24,y=0.36x=0.24,y=0.36
mCu=0.24⋅64=15.36(g)mCu=0.24⋅64=15.36(g)
=> B