Question

Identify the percent yield when 28.16 g of CO2 are formed from the reaction of 456.93 g of C8H18 with 256.0 g of O2.
2C8H18 + 25O2 → 16CO2 + 18H2O

Answer 1

This is a limiting reagent problem (LR). The LR must be identified first
2C8H18 + 25O2 → 16CO2 + 18H2O

1a. mols C8H18 = g/molar mass = 456.93/114 = approx 4, Check that.
1b. mols O2 = 256/32 = approx 8. Check that.
2a.. From 1a, how many mols CO2 can be formed. That's approx 4 mols C8H18 x (16 mols CO2/2 mols C8H18) = approx 4*16/2 = about 32 mols or in grams that's 32 x 44 gCO2/mol = about 1400. Check that number.
2b. From 1b, how many mols CO2 can be formed. That's approox 8 mols O2 x (16 mols CO2/25 mols O2) = 8*16/25 = about 5.12 moles and that converted to g CO2 is 5.12 x 44 g CO3/mol = 225.Check that number.

The LR is the material (C8H18 or O2) which produces the LEAST amount of CO2.. That's O2 in this case so the theoretical yield (TY) is approx 225 g .CO2 The actual yield (AY) from the problem is 28.16 g.
%yield = (AY/TY)*100 = ?

Check all of these calculations. It's late and my eyes are squinting.
Post your work if you get stuck.