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Consider the balanced chemical equation for the combustion of methane (CH4).


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Given that the molar mass of CO2 is 44.01 g/mol, how many liters of oxygen is required at STP to produce 88.0 g of CO2 from this reaction?

44.8 L

45.00 L

89.55 L

89.6 L

Verified Answer :

Took a guess and got it right!

D) 89.6 L

transitionstate

Answer:

                  89.6 L of O₂

Solution:

The balanced chemical equation is as,

                                CH₄  +  2 O₂    →    CO₂  +  2 H₂O

As at STP, one mole of any gas (Ideal gas) occupies exactly 22.4 L of Volume. Therefore, According to equation,

             44 g ( 1 mol) CO₂ is produced by  =  44.8 L (2 mol) of O₂

So,

                  88 g CO₂ will be produced by  = X L of O₂

Solving for X,

                        X = (88 g × 44.8 L) ÷ 44 g

                        X = 89.6 L of O₂

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