Question #f1d6d

Actually, you don't need water to burn methane, you need oxygen gas. This is shown by the balanced chemical equation, which shows oxygen gas as a reactant and water as a product.

#"CH"_ (4(g)) + 2"O"_ (2(g)) -> "CO"_ (2(g)) + 2"H"_ 2"O"_ ((l))#

As you can see, the reaction consumes methane and oxygen gas and produces water. More specifically, for every mole of methane that takes part in the reaction, you get #2# moles of water, assuming, of course, that oxygen gas is in excess.

Use the molar mass of methane to convert the sample to moles

#13.0 color(red)(cancel(color(black)("g"))) * "1 mole CH"_4/(16.04color(red)(cancel(color(black)("g")))) = "0.8105 moles CH"_4#

Since the problem doesn't mention the amount of oxygen gas available for the reaction, you can safely assume that oxygen gas is in excess. This basically means that all the moles of methane will react.

Consequently, you can say that the reaction will produce

#0.8105 color(red)(cancel(color(black)("moles CH"_4))) * ("2 moles H"_2"O")/(1color(red)(cancel(color(black)("mole CH"_4)))) = "1.621 moles H"_2"O"#

Finally, to convert the number of moles of water to grams, you can use the molar mass of water.

#1.621 color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = color(darkgreen)(ul(color(black)("29.2 g")))#

The answer is rounded to three sig figs, the number of sig figs you have for the mass of methane.