How many grams of carbon dioxide will form if 4.4 g of propane reacts with 16.0 g of oxygen producing 7.2 g of water?

Answer 1

This is a limiting reagent. I know that because amounts are given for BOTH reactants.
C3H8 + 5O2 ==> 3CO2 + 4H2O

mols C3H8 = 4.4/molar mass = about 0.1.
mols O2 = 16/molar mass = about 0.5

Convert each to mols H2O, then to grams H2O.
0.1 mol C3H8 x (4 mols H2O/1 mol C3H8) = 0.4 mol H2O.
g = mols x molar mass = about 0.4 x 18 = 7.2 g which is the amount of water produced. This means we guessed right and propane is the limiting reagent.

Now convert mols C3H8 to mols CO2 then to g CO2.
0.1 mol C3H8 x (3 mols CO2/1 mol C3H8) = 0.1 x 3 = 0.3 mol CO2.
g CO2 = mols x molar mass.
Check all of those estimated numbers.