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.