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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.

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  • $\begingroup$ Well, how many moles of water have been produced? $\endgroup$ – Bob Jun 21 '17 at 3:48
  • $\begingroup$ Welcome to SE! feel free to take a tour of this site. $\endgroup$ – Pritt says Reinstate Monica Jun 21 '17 at 4:48
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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 $\ce{H2O}$.

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