Question

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?

Answer 1

`"709 g CH"_4`

Explanation:

Start by writing the balanced chemical equation that describes this reaction

`color(blue)(2)"C"_ ((s)) + 2"H"_ 2"O"_ ((g)) -> "CO"_ (2(g)) + "CH"_ (4(g))`

Notice that the reaction produces `1` mole of methane for every `color(blue)(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 `"16.04 g"` of methane for every `"24.022 g"` of carbon that react.

This means that `"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 `"100 g"` of methane that could be produced, you only get `"85.0 g"`.

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

`834.65 color(red)(cancel(color(black)("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.