# Coal gasification is a process that converts coal into methane gas. If this reaction has a percent yield of 85.0%, how much methane can be obtained from 1250 g of carbon?

Nov 28, 2016

${\text{709 g CH}}_{4}$

#### Explanation:

Start by writing the balanced chemical equation that describes this reaction

$\textcolor{b l u e}{2} {\text{C"_ ((s)) + 2"H"_ 2"O"_ ((g)) -> "CO"_ (2(g)) + "CH}}_{4 \left(g\right)}$

Notice that the reaction produces $1$ mole of methane for every $\textcolor{b l u e}{2}$ moles of carbon that take part in the reaction.

This represents the reaction's theoretical yield, i.e. what you get if the reaction has a 100% yield.

You can convert this mole ratio to a gram ratio by using the molar masses of carbon and methane

2 color(red)(cancel(color(black)("moles C"))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = "24.022 g"

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

You can thus say that the theoretical yield of the reaction will have it produce $\text{16.04 g}$ of methane for every $\text{24.022 g}$ of carbon that react.

This means that $\text{1250 g}$ of carbon will theoretically produce

1250 color(red)(cancel(color(black)("g C"))) * "16.04 g CH"_4/(24.022color(red)(cancel(color(black)("g C")))) = "834.65 g CH"_4

Now, the reaction is said to have an 85.0% yield, which basically means that for every $\text{100 g}$ of methane that could be produced, you only get $\text{85.0 g}$.

As a result, the actual yield of the reaction will be

$834.65 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g CH"_4))) * "85.0 g produced"/(100color(red)(cancel(color(black)("g CH"_4)))) = color(darkgreen)(ul(color(black)("709 g CH}}_{4}}}}$

The answer is rounded to three sig figs.