Question

How many grams of O2(g) are needed to completely burn 48.5 g of C3H8(g)?

Answer 1

Answer:

176g

Explanation:

To appropriately answer this question, we will need to write a complete chemical equation . Firstly what do we know? We know that the molecule given in the question is a hydrocarbon and it burns in oxygen to form carbon iv oxide and water. The reaction equation is written below:

C3H8(g)+5O2(g) --> 3CO2(g)+4H2O(g)

From the equation, we can see that one mole of propane needed 5 moles of oxygen for complete combustion. This is the theoretical relation. Let us find the actual number of moles of propane that burned.

To calculate this, what we simply do is to disburse the mass of propane by the molar mass of propane. The molar mass of propane here is 3(12) + 8(1) = 36 + 8 = 44g/mol

The number of moles of propane is thus = 48.5/44 = 1.102 moles.

Now since one mole of propane needed 5 moles of oxygen , this means 1.102 moles of propane will need 5 * 1.102 moles of oxygen = 5.5 moles

We then proceed to calculate the mass of oxygen gas needed.

The mass of oxygen gas needed equals the number of moles multiplied by the molecular mass of the oxygen molecule. The molecular mass of the oxygen molecule is 2(16) = 32g/mol

The mass needed is thus 5.5 * 32 = 176g

Answer 2

Answer:

g O2(g) = 175.964 g

Explanation:

• C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O

                               3 - C - 3

                               10 - O - 10

                               8 - H - 8

∴ Mw C3H8(g) = 44.1 g/mol

⇒ n C3H8(g) = (48.5 g C3H8(g))×(mol C3H8/44.1 g C3H8) = 1.0997 mol C3H8

⇒ n O2(g) = (1.0997 mol C3H8)×( 5 mol O2(g)/mol C3H8(g) ) = 5.4988 mol O2(g)

∴ Mw O2(g) ≅ 32 g/mol

⇒ g O2(g) = (5.4988 mol O2(g))×( 32 g O2(g) /mol ) = 175.964 g O2(g)