Answer:The first thing you need to do is to write the correctly balanced equation for the combustion of C2H2.
2C2H2 + 5O2 ==> 4CO2 + 2H2O ... balanced equation
Next, we find which reactant, if any, is limiting. There are several ways to do this, but one easy way is to find moles of each reactant, and divide that value by the coefficient in the balanced equation and see which is less.
For C2H2, we have...
1.26 g x 1 mole/26.04 g = 0.0484 moles C2H2 (÷2 = 0.0242)
For O2, we have...
3.39 g O2 x 1 mol/32 g = 0.1059 moles O2 (÷5 = 0.0211) <- This is less than 0.0242 so it is limiting
Next, we use the moles of the limiting reactant (0.1059, not 0.0211) to find moles and then grams of CO2:
0.1059 moles O2 x 4 moles CO2/5 mole O2 x 44 g CO2/mole CO2 = 3.73 g CO2
Explanation:
