PTHH: 2Al +3CuSO4→Al2(SO4)3+3Cu
-Gọi nAl là x ⇒mAl= 27x (gam)
-Theo PTHH ta có:
nCu = \(\frac{3}{2}\)nAl = \(\frac{3}{2}\)x⇒ mCu=96x (gam)
mtăng =2,07 (gam)
⇒96x -27x = 2,07
⇒69x=2,07
⇒x=0,03(mol)
+mAl=0,03.27=0,81(gam)
+nCu=\(\frac{3}{2}n_{Al}=0,045\left(mol\right)\)
+mCu=0,045.64=2,88(gam)