a,
Ta có :
\(m_{Al}=49,1\%.22=10,8g\)
->nAl=0.4
\(\Rightarrow m_{Fe}=22=10,8=11,2\text{->nFe=0.2}\)
\(\text{Fe+2HCl-->FeCl2+H2}\)
\(\text{2Al+3HCl-->2AlCL3+3H2}\)
Theo phương trình:
\(\Rightarrow\text{nHCl=2nFe+3nAl=1,6 mol}\)
b,
\(\Rightarrow\text{nH2=nFe+1,5nAl=0,8mol}\)
->V=17.92l
c,\(n_{CuO}=\frac{72}{80}=0,9\left(mol\right)\)
PTHH:\(\text{ H2+CuO-->H2O+Cu}\)
Trước : 0.8......0.9
Phứng :0.8......0.8
Sau :.....0.........0.1.........0.8.......0.8
\(\Rightarrow\text{m rắn=0,8.64+0,1.80=59,2g}\)
