\(Fe2O3+3H2-->2Fe+3H2O\)
\(n_{Fe2O3}=\frac{12}{160}=0,075\left(mol\right)\)
\(n_{H2}=3n_{Fe}=0,225\left(mol\right)\)
\(V_{H2}=0,225.22,4=5,04\left(l\right)\)
b)\(S+O2-->SO2\)
\(n_O=n_S=1,5\left(mol\right)\)
Số nguyên tử O2=\(1,5.6.10^{23}=9.10^{23}\)