\(n_{H_2}=\dfrac{112}{22,4}=5\left(mol\right)\)
Bảo toàn nguyên tố H : \(n_{H_2}.2=n_{H_2O}.2\)
=> nH2O =5(mol)
=> m H2O = 5.18= 90(g)
\(n_{H_2}=\dfrac{112}{22.4}=5\left(mol\right)\)
\(2H_2+O_2\underrightarrow{^{^{t^0}}}2H_2O\)
\(5...............5\)
\(m_{H_2O}=5\cdot18=90\left(g\right)\)