The equation below represents combustion of methane ($\ce{CH4}$, $\pu{16.04 g/mol}$). Balance the equation, and calculate the mass of water ($\pu{18.02 g/mol}$) formed when $\pu{40.0g}$ of methane is burned. $$\ce{CH4(g) + O2(g) -> CO2 + H2O(g)}$$
I have started the problem by balancing:
$$\ce{1CH4(g) + 2O2(g) -> 1CO2 + 2H2O}$$
I then continued by calculating the amount of substance of methane:
$$\ce{C ($\pu{12g/mol}$) + H ($\pu{4 \times 1g/mol}$) = $\pu{16 g/mol}$ \implies ($\pu{40 g}$ methane) / ($\pu{16 g/mol}$) = $\pu{2.5 mol}$}$$
From here I wold like to know how to convert or get to amount of substance of $\ce{H2O}$ to get the mass of $\ce{H2O}$ produced in this reaction.