A sample of 5.31 g of liquid 1‑propanol, C3H8O, is combusted with 46.4 g of oxygen gas. Carbon dioxide and water are the products.?

Molar mass of C₃H₈O (1- propanol) = (12.0×3 + 1.0×8 + 16.) g/mol = 60.0 g/mol
Molar mass of O₂ (oxygen) = 16.0×2 g/mol = 32.0 g/mol
Molar mass of CO₂ (carbon dioxide) = (12.0 + 16.0×2) = 44.0 g/mol

Initial number of moles of C₃H₈O = (5.31 g) / (60 g/mol) = 0.0885 mol
Initial number of moles of O₂ = (46.4 g) / (32.0 g/mol) = 1.45 mol

2C₃H₈O + 9O₂ → 6CO₂ + 8H₂O
Molar ratio C₃H₈O : O₂ : CO₂ = 2 : 9 : 6

If C₃H₈O completely combusts, O₂ needed = (0.0885 mol) × (9/2) = 0.398 mol < 1.45 mol
Hence, O₂ is in excess, and C₃H₈O (1-propanol) is the limiting reactant.
Number of moles of C₃H₈O reacted = 0.0885 mol

Number of moles of CO₂ produced = (0.0885 mol) × (6/2) = 0.2655 mol
Mass of CO₂ produced = (0.2655 mol) × (44.0 g/mol) = 11.7 g

Number of moles of excess O₂ reacted = 0.398 mol
Mass of excess O₂ remained = [(1.45 - 0.398) mol] × (32 g/mol) = 33.7 g

Combustion .....

Only a gas undergoes combustion. When solving a limiting reactant problem you can compute the amount of product using each reactant. The lesser amount is the theoretical yield and tells the limiting reactant.

2C3H7OH(g) + 9O2(g) --> 6CO2(g) + 8H2O(g)
5.31g ............... 46.4g ...........?g

5.31g C3H7OH x (1 mol C3H7OH / 60.09g C3H7OH) x (3 mol CO2 / 1 mol C3H7OH) x (44.0g CO2 / 1 mol CO2) = 11.7g CO2
46.4g O2 x (1 mol O2 / 32.0g O2) x (2 mol CO2 / 3 mol O2) x (44.0g CO2 / 1 mol CO2) = 42.5g CO2
A mass of 11.7g of CO2 is produced, and propanol is the limiting reactant.
5.31g C3H7OH x (1 mol C3H7OH / 60.09g C3H7OH) x (9 mol O2 / 2 mol C3H7OH) x (32.0g O2 / 1 mol O2) = 12.7g O2 .... reacts

Excess O2 = 46.4g - 12.7g = 33.7g O2