\(n_{Pư}=\frac{376-280}{16.3}=2\)
\(n_{ban.dau}=\frac{376}{122,5.\left(100-2,26\right)}=3\)
\(\Rightarrow H=\frac{2}{3}.100\%=66,67\%\)
\(n_{H2}=\frac{156,8}{22,4}=7\left(mol\right)\)
\(2KClO_3\underrightarrow{^{to,MnO2}}2KCl+3O_2\)
2_____________________3__(mol)
\(H_2+\frac{1}{2}O_2\rightarrow H_2O\)
6____3__________(mol)
\(n_{H2\left(dư\right)}=7-6=1\left(mol\right)\)
\(\Rightarrow m_{H2\left(Dư\right)}=1.2=2\left(g\right)\)
\(\Rightarrow m_{H2O}=6.18=108\left(g\right)\)