nH2= 6.72/22.4=0.3 mol
2K + 2H2O --> 2KOH + H2
0.6____________0.6___0.3
mK=m=0.6*39=23.4g
mdd sau pư= 234+400-0.6=422.6g
C%KOH= 33.6/422.6*100%=7.95%
2K + 2H2O → 2KOH + H2
\(n_{H_2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
a) Theo pT: \(n_K=2n_{H_2}=2\times0,3=0,6\left(mol\right)\)
\(\Rightarrow m=m_K=0,6\times39=23,4\left(g\right)\)
b) Theo PT: \(n_{KOH}=n_K=0,6\left(mol\right)\)
\(\Rightarrow m_{KOH}=0,6\times56=33,6\left(g\right)\)
\(m_{H_2}=0,3\times2=0,6\left(g\right)\)
Ta có: \(m_{dd}saupư=23,4+400-0,6=422,8\left(g\right)\)
\(\Rightarrow C\%_{KOH}=\frac{33,6}{422,8}\times100\%=7,95\%\%\)
