2K + 2H2O → 2KOH + H2
Mg + H2O → X
\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
Theo PT: \(n_K=2n_{H_2}=2\times0,05=0,1\left(mol\right)\)
\(\Rightarrow m_K=0,1\times39=3,9\left(g\right)\)
\(\Rightarrow m_{Mg}=7,1-3,9=3,2\left(g\right)\)
\(\Rightarrow\%m_K=\dfrac{3,9}{7,1}\times100\%=54,93\%\)
\(\Rightarrow\%m_{Mg}=100\%-54,93\%=45,07\%\)