Question

Question #3d9fc

Answer 1

(assuming butane is burnt in excess oxygen) you need to convert grams of butane to mols using its molar mass, use the mol ratio of butane and water to find how many mols of water you will produce, then convert mols of water to grams using molar mass.

Explanation:

`2"C"_4"H"_10(g) + 13"O"_2(g) -> 8"CO"_2(g) + 10"H"_2"O"(l)`

So

`"0.500 g butane" xx ("1 mol butane"/"58.14 g") xx ("10 mols water"/"2 mols butane") xx ("18.02 g" /"1 mol water") = "answer"`