Answer:
       66 g of CO₂
Solution:
The Balance Chemical Reaction is as follow,
               C₂H₂  +  5/2 O₂   →   2 CO₂  +  H₂O
Or,
               2 C₂H₂  +  5 O₂   →   4 CO₂  +  2 H₂O   -------  (1)
Step 1: Find out the limiting reagent as;
According to Equation 1,
      56.1 g (2 mole) C₂H₂ reacts with  =  160 g (5 moles) of O₂
So,
         125 g of C₂H₂ will react with  =  X g of O₂
Solving for X,
           X =  (125 g × 160 g) ÷ 56.1 g
           X =  356.5 g of O₂
It means for total combustion of Ethylene we require 356.5 g of Oâ‚‚, but we are only provided with 60.0 g of Oâ‚‚. Therefore, Oâ‚‚ is the limiting reagent and will control the yield.
Step 2: Calculate Amount of COâ‚‚ produced as;
According to Equation 1,
       160 g (5 mole) O₂ produces  =  176 g (4 moles) of CO₂
So,
         60.0 g of O₂ will produce  =  X g of CO₂
Solving for X,
           X =  (60.0 g × 176 g) ÷ 160 g
           X =  66 g of CO₂
