What are the volumes of each gas in the resulting mixture after the combustion of #"20 cm"^3 color(white)(l)"CO"# with #"30 cm"^3color(white)(l)"O"_2#?

1 Answer
Apr 13, 2017

Answer:

The resulting gas consists of #"20 cm"^3 color(white)(l) "CO"_2# and #"20 cm"^3 color(white)(l) "O"_2#.

Explanation:

All the substances are gases, so we can use Gay-Lussac's Law of Combining Volumes:

The ratio of the volumes in which gases react and form products is the same as the ratio of their moles.

The equation for the reaction is

#"Theor. Vol./cm"^3: color(white)(ml)2color(white)(mmll)1color(white)(mmm)2#
#color(white)(mmmmmmmmm)"2CO" + "O"_2 → "2CO"_2#
#"Actual Vol./cm"^3: color(white)(m)20color(white)(mm)30#
#"Divide by:" color(white)(mmmmml)2color(white)(mmll)1#
#"Moles of reaction:" color(white)(ml)10 color(white)(mm) 30#

The coefficients of #"CO"# and #"O"_2# in the equation tell us that the molecules react in a 2:1 ratio.

A quick way to identify a limiting reactant is to calculate its "moles of reaction".

We divide the number of moles (here they are #"cm"^3#) by their coefficients in the balanced equation.

I did that for you in the lines above.

We see that #"CO"# is the limiting reactant, because it gives the fewest moles of reaction.

Calculate volume of #"CO"_2# formed

#"Vol. of CO"_2 = 20 color(red)(cancel(color(black)("cm"^3color(white)(l) "CO"))) × ("2 cm"^3 color(white)(l)"CO"_2)/(2 color(red)(cancel(color(black)("cm"^3 color(white)(l)"CO")))) = "20 cm"^3color(white)(l) "CO"#

Calculate volume of #"O"_2# reacted

#"Vol of O"_2 color(white)(l)"reacted" = 20 color(red)(cancel(color(black)("cm"^3color(white)(l) "CO"))) × ("1 cm"^3color(white)(l) "O"_2)/(2 color(red)(cancel(color(black)("cm"^3 color(white)(l)"CO")))) = "10 cm"^3color(white)(l) "O"_2#

Calculate the volume of #"O"_2# remaining

#"O"_2color(white)(l) "remaining" = "30 cm"^3 - "10 cm"^3 = "20 cm"^3#