When 125.0 g of ethylene (c2h4) burns in 60.0 grams of oxygen to give carbon dioxide and water, how many grams of co2 are formed? (hint: balance the equation and determine limiting reactant first)

             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 O₂, but we are only provided with 60.0 g of O₂. Therefore, O₂ is the limiting reagent and will control the yield.

Step 2: Calculate Amount of CO₂ 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

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