# 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?

## ${\text{CH"_4 + "H"_2"O" -> "CO" + 3"H}}_{2}$ What volume of ${\text{H}}_{2}$ can be obtained from ${\text{100 cm}}^{3}$ of methane at STP? a)100 cm³ b) 150cm³ c) 300cm³ d)200cm³

Aug 6, 2016

${\text{300 cm}}^{3}$

#### 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.

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{n}_{1} / {n}_{2} = {V}_{1} / {V}_{2}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ the mole ratio is equivalen to the volume ratio

The balanced chemical equation that describes your reaction looks like this

${\text{CH"_ (4(g)) + "H"_ 2"O"_ ((g)) -> "CO"_ ((g)) + color(red)(3)"H}}_{2 \left(g\right)}$

Notice that every mole of methane that takes part in the reaction produced $\textcolor{red}{3}$ 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 $1 : \textcolor{red}{3}$ mole ratio that exists between the two chemical species in the balanced chemical equation will be equivalent to a $1 : \textcolor{red}{3}$ 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 ${\text{22.7 L mol}}^{- 1}$ for STP conditions defined as a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$.

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)|)))