Question

What mass of water would result given complete combustion of a `44.0*g` mass of propane?

Answer 1

Approx. `72*g`.

Explanation:

`C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O`

`"Moles of propane"` `=` `(44.0*g)/(44.10*g*mol^-1)` `~=` `1*mol`. Given the balanced equation, how many moles of water will result, and how much mass does this constitute?