A10 gr mixture of C2H6 and C3H4 gases are combusted?

Molar mass of C₂H₆ = (12.0×2 + 1.0×6) g/mol = 30.0 g/mol
Molar mass of C₃H₄ = (12.0×3 + 1.0×4) g/mol = 40.0 g/mol
Molar mass of CO₂ = (12.0 + 16.0×2) g/mol = 44.0 g/mol

Let y g be the mass of C₂H₆ in the mixture.
Then, mass of C₃H₄ in the mixture = (10 - y) g

Consider the combustion of C₂H₆ :
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Molar ratio C₂H₆ : CO₂ = 2 : 4 = 1 : 2
No. of moles of C₂H₆ burned = (y g) / (30.0 g/mol) = y/30 mol
No. of moles of CO₂ gas produced in the combustion of C₂H₆ = (y/30 mol) × 2 = y/15 mol
Mass of CO₂ gas produced in the combustion of C₂H₆ = (y/15 mol) × (44.0 g/mol) = 44y/15 g

Consider the combustion of C₃H₄ :
C₃H₄ + 4O₂ → 3CO₂ + 2H₂O
Molar ratio C₃H₄ : CO₂ = 1 : 3
No. of moles of C₃H₄ burned = [(10 - y) g] / (40.0 g/mol) = (10 - y)/40 mol
No. of moles of CO₂ gas produced in the combustion of C₃H₄ = [(10 - y)/40 mol] × 3 = (30 - 3y)/40 mol
Mass of CO₂ gas produced in the combustion of C₃H₄ = [(30 - 3y)/40 mol] × (44.0 g/mol) = (33 - 3.3y) g

Total mass of CO₂ gas produced = 30.8 g
(44y/15) + (33 - 3.3y) = 30.8
[(44y/15) + (33 - 3.3y)] × 15 = 30.8 × 15
44y + 495 - 49.5y = 462
5.5y = 33
y = 6

Mass of C₂H₆ in the mixture = 6 g

Balanced equations for the two reactions:
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)
C3H4(g) + 4O2(g) → 3CO2(g) + 2H2O(l)
Let the mass of C2H2 = Xg
The mass of C3H4 = (10 - X)g
Molar mass C2H6 = 30g/mol
Molar mass C3H4 = 40g/mol
Molar mass CO2 = 44g/mol

Mol C2H6 combusted = X/30 mol
This will produce 2X/30 mol CO2
Mass of CO2 produced = 2X/30*44 = 2.93Xg CO2

Mol C3H4 combusted = (10-X) / 40
This will produce 3(10-X) / 40 mol CO2 = (30 - 3X) / 40 Mol CO2
Mass CO2 produced = (30-3X) *44/40
Mass of CO2 produced = (33 -3.3X)g CO2
Total CO2 produced = 30.8g
2.93X + 33 - 3.3X = 30.8
2.93X - 3.3X = 30.8-33
0.37X = 2.2
X = 5.95g C2H6 combusted
Mass of C2H6 = 5.95 g
Mass of C3H4 = 4.05g

Let us check:
5.95g C2H6 = 5.95 / 30 = 0.198 mol
This will produce 0.198*2 = 0.397 mol CO2 = 0.397*44 = 17.45g CO2
4.05g C3H4 = 4.05/40 = 0.101 mol .
This will produce 0.303 mol CO2 0.303*44 = 13.37 g CO2
Total mass CO2 = 17.45+13.37 = 30.8g CO2 produced
Our check confirms answer correct

2C2H6+7O2=4CO2+6H2O
C3H4+4O2=3CO2+2H2O
the total co2 is 30.8 so we can take one of them as x and one of them as 30.8-x
Then find mole of co2 then according to mole of carbon dioxide you can find mole of C2H6 and C3H4 then multiply them by molar mass to find mass of them then add them which is equal to ten put them in equation and find x which is 13.2 then from x value find mass of C3H4 which is 6gr