Robert S. answered • 04/10/21

Patient PhD Chemist with 40 Years of R&D And Teaching

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.

C_{4}H_{6} + O_{2} 🡪 CO_{2} + H_{2}O

If we start with 1 C_{4}H_{6} , we see the following:

1C_{4}H_{6} + ?O_{2} 🡪 4CO_{2} + 3H_{2}O

I found homes for all 4 carbons and 6 hydrogens by using coefficients of 4 and 3 for CO_{2} and H_{2}O. 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 O_{2}. As described in another answer, we can side with the mathematicians for a moment and assign a coefficient of 5.5 to the O_{2} to get 11 oxygen atoms.

1C_{4}H_{6} + 5.5O_{2} 🡪 4CO_{2} + 3H_{2}O

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

2C_{4}H_{6} + 11O_{2} 🡪 8CO_{2} + 6H_{2}O

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 C_{4}H_{6}, 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