Question

How do you calculate the amount of ethene (in moles) in 100 cm3?

Answer 1

The limiting reactant is ethene.

Explanation:

You don’t have to convert volumes to moles. You can use volumes, because moles are directly proportional to volumes.

We can use Gay-Lussac's Law of Combining volumes to solve this problem.

We know that we will need a balanced equation with volumes and molar masses of the compounds involved.

1. Gather all the information in one place with volumes below the formulas.

`"Theor./cm"^3:color(white)(mm)1color(white)(mmmmm)2color(white)(mmmmmll)2`
`color(white)(mmmmmmml)"C"_2"H"_4 color(white)(m)+color(white)(m) "2O"_2color(white)(m) → color(white)(m)"2CO"_2 + "2H"_2"O"`
`"Vol./cm"^3:color(white)(mml)100color(white)(mmmm)400`
`"Divide by:"color(white)(mmml)1color(white)(mmmmm)2`
`"Moles rxn":color(white)(mml)100color(white)(mmmlll)200`

2. Identify the limiting reactant

An easy way to identify the limiting reactant is to calculate the moles of reaction each will give.

Since the volume is directly proportional to the number of moles. the volume ratio is the same as the molar ratio

In this case, you can divide the volume of each reactant by its corresponding coefficient in the balanced equation.

I did that for you in the table above.

Ethene is the limiting reactant because it gives the fewest "moles of reaction".

3. Calculate the theoretical yield of `"CO"_2` from each reactant

If you don’t believe me, here's the calculation.

`"From C"_2"H"_4 = 100 color(red)(cancel(color(black)("cm"^3color(white)(l) "C"_2"H"_4))) × ("2 cm"^3color(white)(l) "CO"_2)/(1 color(red)(cancel(color(black)("cm"^3 color(white)(l)"C"_2"H"_4)))) = "200 cm"^3 color(white)(l)"CO"_2`

`"From O"_2 = 400 color(red)(cancel(color(black)("cm"^3color(white)(l) "O"_2))) × ("2 cm"^3 color(white)(l)"CO"_2)/(2 color(red)(cancel(color(black)("cm"^3color(white)(l) "O"_2)))) = "400 cm"^3color(white)(l) "CO"_2`

`"C"_2"H"_4` is the limiting reactant because it gives the fewest moles of `"CO"_2`.