Final answer:
The volume of air needed for the complete combustion of 50 dm³ of the methane and ethene mixture is 575 dm³, based on oxygen requirements for combustion and the oxygen content in air.
Step-by-step explanation:
To calculate the volume of air needed for the complete combustion of 50 dm³ of a mixture of 70% methane and 30% ethene by volume, we first need to determine the volume of oxygen required for this combustion and then relate that to the volume of air needed based on the oxygen content of air.
The complete combustion of methane (CH4) is represented by the equation:
CH4 + 2 O2 → CO2 + 2 H2O
For ethene (C2H4), the combustion reaction is:
C2H4 + 3 O2 → 2 CO2 + 2 H2O
Thus, one volume of CH4 needs two volumes of O2, and one volume of C2H4 needs three volumes of O2. For the 50 dm³ mixture, 70% is CH4, which is 35 dm³ and requires 70 dm³ of O2, and 30% is C2H4, which is 15 dm³ and requires 45 dm³ of O2. The total oxygen needed is 115 dm³.
Since air contains 20% oxygen by volume, we can now calculate the volume of air needed:
(115 dm³ O2) ÷ (0.20) = 575 dm³ of air
Thus, the correct answer is not listed in the options provided by the student. Therefore, the student or the question may have a mistake as 575 dm³ of air is required for complete combustion.