Question

5) At STP, 33.6 L of hydrogen gas is used to saturate 24 grams of a `"C"_2"H"_4` and `"C"_2"H"_2` mixture. What is the mass percentage of `"C"_2"H"_4` in the mixture? (C:12, H:1)

Answer 1

Warning! Long Answer. 5) The mixture contains 37 % by mass `"C"_2"H"_4`.

Explanation:

The equations for the hydrogenation reactions are

`"C"_2"H"_4 + "H"_2 → "C"_2"H"_6`
`"C"_2"H"_2 + "2H"_2 → "C"_2"H"_6`

STP is 1 bar and 0 °C. Under these conditions, the molar volume of an ideal gas is 22.71 L.

`"Moles of H"_2 = 33.6 color(red)(cancel(color(black)("L"))) × "1 mol"/(22.71 color(red)(cancel(color(black)("L")))) = "1.48 mol"`

Two conditions must be satisfied:

`"Mass of C"_2"H"_4 + "mass of C"_2"H"_2 = "total mass"`
`"Moles of H"_2 color(white)(l)"for C"_2"H"_4 + "moles of H"_2 color(white)(l)"for C"_2"H"_2 = "total moles of H"_2`

Let `x = "mass of C"_2"H"_4` and `y = "mass of C"_2"H"_2"`

Then

`bb((1)) color(white)(m)x + y = 24`

`"Moles of C"_2"H"_4 = x color(red)(cancel(color(black)("g C"_2"H"_4))) × ("1 mol C"_2"H"_4)/(28 color(red)(cancel(color(black)("g C"_2"H"_4)))) = 0.0357x color(white)(l)"mol C"_2"H"_4`

`"Moles of H"_2color(white)(l) "for C"_2"H"_4 = 0.0357x color(red)(cancel(color(black)("mol C"_2"H"_4))) × ("1 mol H"_2)/(1 color(red)(cancel(color(black)("mol C"_2"H"_4)))) = 0.0357x color(white)(l)"mol H"_2`

`"Moles of C"_2"H"_2 = y color(white)(l)color(red)(cancel(color(black)("g C"_2"H"_2))) × (1 "mol C"_2"H"_2)/(26 color(red)(cancel(color(black)("g C"_2"H"_2)))) = 0.0385ycolor(white)(l) "mol C"_2"H"_2`

`"Moles of H"_2color(white)(l) "for C"_2"H"_2 = 0.0385y color(red)(cancel(color(black)("mol C"_2"H"_2))) × ("2 mol H"_2)/(1 color(red)(cancel(color(black)("mol C"_2"H"_2)))) = 0.0769y color(white)(l)"mol H"_2`

`bb((2))color(white)(m)0.0357 x + 0.0769y = 1.48`

From Equation 1,

`bb((3))color(white)(m)y = 24 - x`

Substitute Equation 3 into Equation 2.

`0.0357x + 0.0769(24-x) = 1.48`

`0.0357x + 1.846 - 0.0769x = 1.48`

`0.0412x = 0.366`

`x = 0.366/0.0412 = 8.9`

`y = 24 - x = 24 - 8.9 = 15.1`

`"Mass of C"_2"H"_4 = "8.9 g"`
`"Mass of C"_2"H"_2 = "15.1 g"`

`"Mass % C"_2"H"_4 = (8.9 color(red)(cancel(color(black)("g"))))/(24 color(red)(cancel(color(black)("g")))) × 100 % = 37 %`

Check:

`8.9/28 + 2 ×15.1/26 = 1.48`

It checks!