How many grams of O2(g) are needed to completely burn 24.0 g of C3H8(g)?

Answer 1

Here are the steps to work any simple stoichiometry problem (like this one). You must make a slight adjustment when you get to limiting reagent problems but they are almost the same. Print this out. Keep it. Memorize the steps.

Step 1. Write and balance the equation.
C3H8 + 5O2 ==> 3CO2 + 4H2O

Step 2. Convert grams of what you have (in this case g C3H8) to mols. mols C3H8 = grams/molar mass = 24.0/44 = 0.545

Step 3. Using the coefficient in the balanced equation, convert mols of what you have (C3H8) to mols of what you want (in this case mols O2).
0.545 mols C3H8 x (5 mols O2/1 mol C3H8 = 0.545 x (5/1) = 2.73

Step 4. Convert mols of what you want (O2) to grams. g = mols x molar mass
g = 2.727 x 32 = ? and round to 3 significant figures.

Answer 2

Balance the equation
C3H8 + _n_ O2 = 3 CO2 + 4 H2O

mass(O2) = n * mass(C3H8) * molarmass(O2) / molarmass(C3H8)