Calculate the mass of carbon dioxide produced from the combustion of 11.2 g of C4H6. (finish the equation and balance) C4H6 + O2 —-> CO2 + H2O

Question

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1. Calculate the mass of carbon dioxide produced from the combustion of 11.2 g of C4H6. (finish the equation and balance) 

             C4H6  +   O2 🡪   CO2 +  H2O

Answer 1

Hello, Lisa,

This is a slightly more difficult question than the others. Balancing the equation takes a little more time. One useful technique to balancing is to assign a coefficient of "1" to the most complex compound. In this case I would pick C4H6, butyne, a triple-bonded organic compound out looking for a good time (highly reactive). It's not terribly complex, but it has the most atoms that need to find homes on the product side.

C₄H₆  +   O₂ 🡪   CO₂ +  H₂O

If we start with 1 C₄H₆ , we see the following:

1C₄H₆  +   ?O₂ 🡪   4CO₂ +  3H₂O

I found homes for all 4 carbons and 6 hydrogens by using coefficients of 4 and 3 for CO₂ and  H₂O. But the oxygen is a problem. We have 11 oxygen atoms on the product side, but oxygen is a diatomic, and only comes in pairs, as O₂. As described in another answer, we can side with the mathematicians for a moment and assign a coefficient of 5.5 to the O₂ to get 11 oxygen atoms.

1C₄H₆  +   5.5O₂ 🡪   4CO₂ +  3H₂O

Before the chemists can reprimand you, multiply all coefficients by 2 to bring the O₂ coefficient to 11, This results in:

2C₄H₆  +   11O₂ 🡪   8CO₂ +  6H₂O

This is now legally balanced.

11.2 grams of butyne with a molar mass of 54 g/mole yields 0.207 moles of butyne.

We get 8 moles of carbon dioxide for every 2 moles of C₄H₆, a molar ratio of 8/2 or 4. That means we'll get 4*(0.207 moles) = 0.829 moles of CO2. With a molar mass of 44 grams/mole, that amounts to 36.5 grams of carbon dioxide.

Bob