How many grams of methane gas (CH4) need to be combusted to produce 18.2 L water vapor at 1.2 atm and 275 K? Show all of the work used to solve this problem.

Start by writing out the balanced chemical equation. You should know that combustion reactions generally follow the addition of oxygen ( that means, reacting methane with oxygen) to produce carbon dioxide and water. Hence, you have: \[CH4 + O_2 \rightarrow CO_2 + H_2O \] I'll leave it to you to balance it. Next, use the ideal gas law: PV = nRT where P = pressure [atm] V = volume [L] n = mole R = gas constant (you can look up the value in your textbook, but i think it's 0.08314???) T = temperature [K] Now, you're being asked to find grams. So this is where your balanced chemical equation will come in handy. Start by finding the moles of given information. (This means, using algebra, rearrange the formula PV=nRT to solve for "n") Once you have "n". Convert moles (n) of water -> moles methane and then you know the relationship (grams/mol). Use dimensional analysis. The end.

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