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

1 Answer
Aug 6, 2016

Answer:

#"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)|)))#