How to calculate mass produced when a given gas is burned? 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. 
  $$\ce{CH4(g) + O2(g) -&gt; CO2 + H2O(g)}$$


I have started the problem by balancing:

$$\ce{1CH4(g) + 2O2(g) -&gt; 1CO2 + 2H2O}$$

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.
You have already balanced the equation, so we can get the stoichiometric  coefficients. From the equation we can say that 1 mole $\ce{CH4}$ gives 2 moles of $\ce{H2O}$ because their coefficients are 1 and 2, correspondigly. So 2.5 moles of $\ce{CH4}$ would give 5 moles of $\ce{H2O}$, that is $\pu{90 g}$ of $\ce{H2O}$.