\(2KClO_3\rightarrow2KCl+3O_2\)
\(Al+3HCl\rightarrow AlCl_3+\frac{3}{2}H_2\)
Ta có: \(n_{KClO3}=\frac{24,5}{39+35,5+16.3}=0,2\left(mol\right)\)
\(\rightarrow n_{O2}=\frac{3}{2}n_{KClO3}=0,3\left(mol\right)\)
\(n_{Al}=\frac{5,4}{27}=0,2\left(mol\right)\rightarrow m_{H2}=\frac{3}{2}n_{Al}=0,3\left(mol\right)\)
\(2H_2+O_2\rightarrow2H_2O\)
Vì nH2 < 2nO2 nên O2 dư
\(\rightarrow\) Hiệu suất tính theo H2
\(\rightarrow n_{H2_{pu}}=0,3.80\%=0,24\left(mol\right)=n_{H2O}\)
\(\rightarrow m_{H2O}=0,24.18=4,32\left(g\right)\)