What are the volumes of each gas in the resulting mixture after the combustion of ${20 cm}^{3} l CO$ $"20 cm"^3 color(white)(l)"CO"$ with ${30 cm}^{3} l O_{2}$ $"30 cm"^3color(white)(l)"O"_2$ ?

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What are the volumes of each gas in the resulting mixture after the combustion of ${20 cm}^{3} l CO$ $"20 cm"^3 color(white)(l)"CO"$ with ${30 cm}^{3} l O_{2}$ $"30 cm"^3color(white)(l)"O"_2$ ?

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Answer 1 · 2017-04-13T20:10:12

The resulting gas consists of ${20 cm}^{3} l {CO}_{2}$ $"20 cm"^3 color(white)(l) "CO"_2$ and ${20 cm}^{3} l O_{2}$ $"20 cm"^3 color(white)(l) "O"_2$ .

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

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

The coefficients of $CO$ $"CO"$ and $O_{2}$ $"O"_2$ 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 ${cm}^{3}$ $"cm"^3$ ) by their coefficients in the balanced equation.

I did that for you in the lines above.

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

Calculate volume of ${CO}_{2}$ $"CO"_2$ 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

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

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What are the volumes of each gas in the resulting mixture after the combustion of ${20 cm}^{3} l CO$ $"20 cm"^3 color(white)(l)"CO"$ with ${30 cm}^{3} l O_{2}$ $"30 cm"^3color(white)(l)"O"_2$ ?

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What are the volumes of each gas in the resulting mixture after the combustion of ${20 cm}^{3} l CO$ $"20 cm"^3 color(white)(l)"CO"$ with ${30 cm}^{3} l O_{2}$ $"30 cm"^3color(white)(l)"O"_2$ ?