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!