\(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)
\(2K+2H_2O\rightarrow2KOH+H_2\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PT: \(n_{H_2O}=2n_{H_2}=0,6\left(mol\right)\)
Theo ĐLBT KL, có: mKL + mH2O = mX + mH2
⇒ mX = 14,2 + 0,6.18 - 0,3.2 = 24,4 (g)