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
What volume of
a)100 cm³ b) 150cm³ c) 300cm³ d)200cm³
1 Answer
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
Assuming that the hydrogen gas is also kept under STP conditions, you can say that the
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
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)|)))#