\(n_{Zn}=\frac{9,75}{65}=0,15\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,15 ------------------------> 0,15 (mol)
=> \(V_{H2}=0,15.22,4=3,36\left(l\right)\)
\(n_{Fe2O3}=\frac{16}{160}=0,1\left(mol\right)\)
PTHH: Fe2O3 + 3H2 --> 2Fe + 3H2
________ 0,1____0,15 (mol)
Xét tỉ lệ: \(\frac{0,1}{1}>\frac{0,15}{3}\) => Fe2O3 dư, H2 hết
PTHH: Fe2O3 + 3H2 --> 2Fe + 3H2O
0,05 <-- 0,15 --> 0,1 --> 0,15 (mol)
=> \(m_{Fe}=0,1.56=5,6\left(g\right)\)
=> \(m_{H2O}=0,15.18=2,7\left(g\right)\)