A limiting Reagent Problem: Check My work

[cheechnchong]

Problem: combustion of propane, C3H8

C3H8 (g) + 5O2 (g) --- 3 CO2 (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 C3H8 of O2?
(b) what mass of CO2 is formed when 1.00 g of C3H8 racts completely?

My Approach:

(a) 5.00g C3H8 x (1mol C3H8/ 44g C3H8) = .114 mol C3H8
10.0g O2 x (1mol C3H8/ 32g O2) = .313 mol O2

C3H8 is limiting

(b) 1.00g C3H8 x (1 mol C3H8/ 44g C3H8) x (3mol CO2/ 1mol C3H8) x (44g CO2/ 1 mol CO2) = 3.00g CO2

I'm kinda doubting my answer on the limiting reagent part...can someone explain it to me if I am wrong?

 

[cheechnchong]

bump anyone?

 

[courtrigrad]

convert to moles first, and then divide by the coefficients. the smaller number indicates which one is the limiting reactant.

So $$O_{2}$$ is the limiting reactant

 

[cheechnchong]

convert to moles first, and then divide by the coefficients. the smaller number indicates which one is the limiting reactant.

So $$O_{2}$$ is the limiting reactant

great tip thanks!
they skip that kinda approach in college...it's all memorization! i hate it...

 

[Borek]

Limiting reagents calculations.