Consider the balanced equation for the combustion of propane, C3H8?

Question

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

If propane reacts with oxygen as above

a. what is the limiting reagent in a mixture containing 5.00 g of C3H8 and 10.0 g of O2?

b. what mass of CO2 is formed when 1.00 g of C3H8 reacts completely?

Answer 1

(a) 5.00 g of C3H8 / 44.1 g/mol = 0.113 moles

0.113 * 5 = 0.565 moles

10.0 g of O2 / 32.0 g/mol = 0.3125 moles, giving Oxygen as limiting reagent.

(b) 1.00 g /44.1 g/mol = 0.0227 moles burned to give

3 * 0.0227 moles = 0.0681 moles of CO2 formed

0.0681 moles * 44.01 g/mole = 3.00 grams of CO2

Answer 2

Propane Combustion Equation

Answer 3

Combustion Of Propane Equation

Answer 4

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RE:
Consider the balanced equation for the combustion of propane, C3H8?
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

If propane reacts with oxygen as above

a. what is the limiting reagent in a mixture containing 5.00 g of C3H8 and 10.0 g of O2?

b. what mass of CO2 is formed when 1.00 g of C3H8 reacts completely?

Answer 5

I'll answer b, first.
b. at mass C=12, at mass O=16
molecular mass CO2 = 12+ 32= 44
=> 1 g C will give 44/12= 3.667g CO2 on complete combustion.

a. What excatly do u mean by limiting reagant?


The limits of combustibility, if that's what u mean, normally ranges between 50% to 150% of Oxygen required .

now, going by the equation,

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
12*3+8 + 5*32
or 44 + 160
ie 44 gm of C3H8 needs 160 g O2 for total combustion
ie 1 g C3 H8 needs 160/44 g O2 for total combustion
or, 5 g C3 H8 needs 160/44 *5 = 18.182 g O2 for total combustion

now, going by combustiblity limits, the reaction will start (O2 available= 10 g, > 50% of 18.182g) and continue until O2 availability is not below 50% as required by balanced equation.

So. here O2 becomes the limiting Reagent

Answer 6

What? Are you guys speaking English or is this just the script for a star trek episode gone bad.