a)
\(n_{H_2O}=\dfrac{45}{18}=2,5\left(mol\right)\)
\(2H_2O\xrightarrow[]{t^o}2H_2+O_2\)
2,5 2,5 1,25 (mol)
\(m_{H_2}=2,5.2=5\left(gam\right);m_{O_2}=1,25.32=40\left(gam\right)\\ \dfrac{m_{H_2}}{m_{O_2}}=\dfrac{5}{40}=\dfrac{1}{8}\)
b)
\(V_{H_2}=2,5.22,4=56\left(l\right);V_{O_2}=1,25.22,4=28\left(l\right)\\ \dfrac{V_{H_2}}{V_{O_2}}=\dfrac{56}{28}=\dfrac{1}{2}\)