\(2H_{2} O _{(l)} -------\ \textgreater \ 2H_{2} _{(g)} + O_{2} _{(g)}\)
moles of oxygen in reaction = \(\frac{mass}{molar mass}\)
= \(\frac{50.00 g}{32.00 g / mol}\)
= 1.563 mol
Mole Ratio of oxygen to water in the reaction = 1 : 2
∴ for every mole of oxygen there is two moles of water
Thus moles of water in the reaction = (moles of oxygen) 2
= 1.563 mol 2
= 3.126 mol
Mass of water needed to produce amount of O = mol molar mass
= (3.126 mol) (18.01 g/mol)
= 56.299 g
≈ 56. 3 g
