https://i.imgur.com/sUtqdkH.png
a)nNO=\(\frac{1,344}{22,4}\)=0,06(mol)
Gọi a là số mol Al b là số mol Cu
Ta có\(\left\{{}\begin{matrix}\text{27a+64b=4,38}\\\text{3a+2b=0,06.3=0,18}\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}\text{a=0,02}\\\text{b=0,06}\end{matrix}\right.\)
mAl=0,02.27=0,54(g)
mCu=0,06.64=3,84(g)
b)
4Al(NO3)3\(\rightarrow\)2Al2O3+12NO2+3O2
2Cu(NO3)\(\rightarrow\)2CuO+4NO2+O2
nAl2O3=\(\frac{0,02}{2}\)=0,01(mol)
nCuO=0,06(mol)
m=0,01.102+0,06.0=5,82(g)
