Question

HELP ME? :(
In an experiment, 44 g of propane, C3H8, was burned, 132 g of carbon dioxide and of 72 g water. The mass of oxygen that was needed for this reaction was ______?

Answer 1

this might help When doing these stoichiometric questions you MUST always use a balanced equation.

C3H8 + 5 O2 --------------> 3 CO2 + 4 H2O and you must know how to READ the equation.

I says: 1 mole of propane combined with 5 moles of O2 produces 3 moles of CO2 and 4 moles of H2O

So, now go to the info in your question. You have 44 g of propane and 132 g of CO2. You must convert both these masses into moles.

moles C3H8 = 44 g / 44.11 = 1.0 mol of propane. Looking at the equation, that whould produce 3 mol of CO2.

Moles of CO2 = 132 g / 44.01 = 3 mol of CO2. So everything checks out. Now all you need to do is find out how much O2 was consumed.

The equation says that for every mol of C3H8 used you need 5 mol of O2. So the number of moles of O2 used is 1 mol x 5 or 5 moles

Now convert that back into a mass.

mass = moles x Molar Mass

mass = 5 moles x 32 g/mol = 160 g of O2

Do you see? You have to use the mol ratios in a balanced equation to work these problems out.

Answer 2

Or this might help. The 132g refers to the mass of CO2 produced - the mass of water is not given. You cannot use the conservation of mass method if you ignore the mass of one product:
Equation: C3H8+ 5O2 → 3CO2 + 4H2O
molar mass propane = 36+8 = 44g/mol
So we are dealing with 1 mol of propane
This will produce 3 mol CO2
Molar mass CO2 = 44g/mol
3mol CO2 = 132g CO2
So that checks out with the data given.

We know from the equation that 5 mol O2 is required to combust the 1 mol propane
Molar mass O2 = 32g/mol
5mol O2 = 5*32 = 160g O2 required
Which also checks out with the given answer.

Answer 3

C 3 H 8+O2-->CO2+H2O

now balance the eq,
C3H8 + 5 O2 = 3 CO2 + 4 H2O

5 O2 =160g