Question

Methane reacts with steam to form `"H"_2` and `"CO"` as shown. What volume of `"H"_2` can be obtained from `"100 cm"^3` of methane at STP?

`"CH"_4 + "H"_2"O" -> "CO" + 3"H"_2`

What volume of `"H"_2` can be obtained from `"100 cm"^3` of methane at STP?

a)100 cm³ b) 150cm³ c) 300cm³ d)200cm³

Answer 1

`"300 cm"^3`

Explanation:

The trick here is to realize that when two gases that take part in a chemical reaction are kept under the same conditions for pressure and temperature, their mole ratio in the balanced chemical equation is equivalent to a volume ratio.

`color(purple)(|bar(ul(color(white)(a/a)color(black)(n_1/n_2 = V_1/V_2)color(white)(a/a)|))) ->` the mole ratio is equivalen to the volume ratio

The balanced chemical equation that describes your reaction looks like this

`"CH"_ (4(g)) + "H"_ 2"O"_ ((g)) -> "CO"_ ((g)) + color(red)(3)"H"_ (2(g))`

Notice that every mole of methane that takes part in the reaction produced `color(red)(3)` moles of hydrogen gas. You know that the methane reacts under STP conditions.

Assuming that the hydrogen gas is also kept under STP conditions, you can say that the `1:color(red)(3)` mole ratio that exists between the two chemical species in the balanced chemical equation will be equivalent to a `1:color(red)(3)` volume ratio.

This means that your sample of methane will produce

`100 color(red)(cancel(color(black)("cm"^3"CH"_4))) * (color(red)(3)color(white)(.)"cm"^3 "H"_2)/(1color(red)(cancel(color(black)("cm"^3"CH"_4)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("300 cm"^3"H"_2)color(white)(a/a)|)))`

You can double-check your answer by using the molar volume of a gas at STP, which is equal to `"22.7 L mol"^(-1)` for STP conditions defined as a pressure of `"100 kPa"` and a temperature of `0^@"C"`.

The molar volume of a gas at STP tells you the volume occupied by one mole of an ideal gas kept under STP conditions. In your case, the volume of methane gas

`100 color(red)(cancel(color(black)("cm"^3))) * (1 color(red)(cancel(color(black)("dm"^3))))/(10^3color(red)(cancel(color(black)("cm"^3)))) * "1 L"/(1color(red)(cancel(color(black)("dm"^3)))) = "0.1 L"`

will contain

`0.1 color(red)(cancel(color(black)("L"))) * "1 mole CH"_4/(22.7color(red)(cancel(color(black)("L")))) = "0.004405 moles CH"_4`

under STP conditions. This means that your reaction will produce

`0.004405 color(red)(cancel(color(black)("moles CH"_4))) * (color(red)(3)color(white)(.)"moles H"_2)/(1color(red)(cancel(color(black)("mole CH"_4)))) = "0.0132 moles H"_2`

If the sample of hydrogen gas is kept under STP conditions, its volume will be equal to

`0.0132 color(red)(cancel(color(black)("moles H"_2))) * "22.7 L"/(1color(red)(cancel(color(black)("mole H"_2)))) = "0.30 L H"_2`

Convert this back to cubic centimeters to get

`0.30 color(red)(cancel(color(black)("L"))) * (1 color(red)(cancel(color(black)("dm"^3))))/(1 color(red)(cancel(color(black)("L")))) * (10^3"cm"^3)/(1color(red)(cancel(color(black)("dm"^3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("300 cm"^3"H"_2)color(white)(a/a)|)))`