Question

A mixture of ethene and propene requires 3 mol of oxygen for complete oxidation and 20.16 L of hydrogen for complete reduction. What is the mass of the original mixture?

Answer 1

Warning! Long Answer. The mass of the initial mixture is 28.1 g.

Explanation:

General method

Usually, we solve this type of problem by setting up two simultaneous equations in two unknowns.

Here, the two equations would be

`bb"(1)"color(white)(m) "Moles of O"_2color(white)(l) "for C"_2"H"_4 + "Moles of O"_2color(white)(l)"for C"_3"H"_6 = "Total moles of O"_2`

`bb"(2)"color(white)(m) "Moles of H"_2color(white)(l) "for C"_2"H"_4 + "Moles of H"_2color(white)(l) "for C"_3"H"_6 = "Total moles of H"_2`

Step 1. Set up Equation (1)

Let `x = "moles of C"_2"H"_4` and `y = "moles of C"_3"H"_6`

The combustion equations are

`bb((a))color(white)(m)"C"_2"H"_4 + "3O"_2 → "2CO"_2 + "2H"_2"O"`
`bb((b))color(white)(m)"2C"_3"H"_6 + "9O"_2 → "6CO"_2 + "6H"_2"O"`

From Equation a: `"Moles of O"_2 = "3 × moles of C"_2"H"_4`

From Equation b: `"Moles of O"_2 = 9/2 × "moles of C"_3"H"_6 = 4.5 × "moles of C"_3"H"_6`

This gives

`bb((1))color(white)(m)3x + 4.5y = 3`

Step 2. Set up Equation (2)

Calculate the moles of `"H"_2`

You don't give the conditions, so I shall assume STP (0 °C and 1 bar).

Under these conditions, the molar volume of an ideal gas is 22.71 L.

∴ `"Moles of H"_2 = 20.16 color(red)(cancel(color(black)("L"))) × "1 mol"/(22.71 color(red)(cancel(color(black)("L")))) = "0.8877 mol"`

The hydrogenation equations are

`bb((c))color(white)(m)"C"_2"H"_4 + "H"_2 → "C"_2"H"_6`
`bb((d))color(white)(m)"C"_3"H"_6 + "H"_2 → "C"_3"H"_8`

From Equation c: `"Moles of H"_2 = "moles of C"_2"H"_4`

From Equation d: `"Moles of H"_2 = "moles of C"_3"H"_6`

This gives

∴ `bb((2))color(white)(m)x + y = 0.8877`

Step 3. Solve the two equations

`bb((1))color(white)(m)3x + 4.5y = 3`
`bb((2))color(white)(m)x + y = 0.8877`

From 2,

`bb((3))color(white)(m)y = 0.8877 -x`

Substitute 3 in 1.

`3x + 4.5(0.8877-x) = 3`

`3x + 3.995 - 4.5x = 3`

`1.5x = 0.995`

`x = 0.995/1.5 = 0.663`

From 2,

`y = 0.8877 -x = 0.8877 - 0.663= 0.225`

`"Moles of C"_2"H"_4= "0.663 mol"`
`"Moles of C"_3"H"_6 = "0.225 mol"`

Step 4, Calculate the mass of the mixture

`"Mass of C"_2"H"_4 = 0.663 color(red)(cancel(color(black)("mol C"_2"H"_4)))× ("28.05 g C"_2"H"_4)/(1 color(red)(cancel(color(black)("mol C"_2"H"_4)))) = "18.6 g C"_2"H"_4`

`"Mass of C"_3"H"_6 = 0.225 color(red)(cancel(color(black)("mol C"_3"H"_6)))× ("42.08 g C"_3"H"_6)/(1 color(red)(cancel(color(black)("mol C"_3"H"_6)))) = "9.47 g C"_3"H"_6`

`"Total mass = 18.6 g + 9.47 g = 28.1 g"`

Check:

`3x + 4.5y = 3 × 0.663 + 4.5 × 0.225 = 1.99 + 1.01 = 3.00`

It checks!