Question

If 12.0 g methane ( ch4 ) reacts completely with oxygen , what is the final yield of water in grams ?

Answer 1

Look at the stoichiometric combustion reaction...

Explanation:

`underbrace(CH_4(g) +2O_2(g))_"16g + 64 g = 80 g" rarr underbrace(CO_2(g) + 2H_2O(l))_"44 g + 36 g = 80 g" + Delta`

All chemical reactions CONSERVE MASS and CHARGE....and clearly this one does as well. With respect to methane, we gots a molar quantity of `(12.0*g)/(16.0*g*mol^-1)=0.75*mol`...and so, given the stoichiometry, `2xx0.75*molxx18.0*g*mol^-1=27.0*g` with respect to water....will result from complete combustion of that methane.

Answer 2

The final yield of water in grams is `"27.0 g"`.

Explanation:

Balanced equation

`"CH"_4("g") + "2O"_2("g")"` `rarr` `"CO"_2("g") + "2H"_2"O(g)"`

Moles methane

Determine moles `"CH"_4"` by dividing its given mass by its molar mass, `"16.043 g/mol"`.

`(12.0color(red)cancel(color(black)("g")) "CH"_4)/(16.043color(red)cancel(color(black)("g"))/"mol")="0.74799 mol CH"_4"`

I am keeping some guard digits to reduce rounding errors. The final answer will be rounded to three significant figures.

Moles water

Determine mol `"H"_2"O"` by multiplying mol `"CH"_4"` by the mol ratio between `"CH"_4"` and `"H"_2"O"` in the balanced equation, with mol `"H"_2"O"` in the numerator.

`0.74799color(red)cancel(color(black)("mol CH"_4))xx(2"mol H"_2"O")/(1color(red)cancel(color(black)("mol CH"_4)))="1.4960 mol H"_2"O"`

Mass water

Determine the mass of `"H"_2"O"` produced by multiplying mol `"H"_2"O"` by its molar mass, `"18.015 g/mol"`.

`1.4960color(red)cancel(color(black)("mol H"_2"O"))xx(18.015"g H"_2"O")/(1color(red)cancel(color(black)("mol H"_2"O")))="27.0 g H"_2"O"` (rounded to three significant figures)