Question

If 13.8 g of butane gas (C4H10) reacts with 21.4 g oxygen gas, what will be the theoretical yield of each product produced in the reaction (assume no side reactions)? Which reactant is in excess and by how much? I want to know the steps because I'm confusing with two way.

Answer 1

Reaction that has been occurred is as follows

C4H10 + 13/2 O2 --- 4CO2 + 5H2O

Since from stoichiometry we can say that

1 mole of C4H10 reacts with = 13/2 moles of O2

Or 58.12 g of C4H10 reacts with = (13 x 32)/2 g of O2

So, 13.8 g of C4H10 reacts with = (13 x 32 x 13.8) / (2 x 58.12)= 49.38 g of O2

but in the question we have only 21.4 g of O2 only

so, here O2 acts as limiting reagent

and C4H10 acts as excess reagent.

So, 21.4 g of O2 reacts with = (58.12 x 2 x 21.4 ) / ( 13 x 32 ) = 5.96 g of C4H10

So, Butane left in the reaction = given mass - used mass = 13.8 - 5.96 = 7.84 g of Butane gets left

As 13/2 moles of O2 give = 4 moles of CO2

or (13 x 32)/2 g of O2 gives = ( 4 x 44) g of CO2

so, 21.4 g of O2 gives = ( 4 x 44 x 21.4 x 2 ) / (13 x 32) = 18.10 g of CO2

also

13/2 moles of O2 gives = 5 moles of H2O

or ( 13 x 32) /2 g of O2 gives = ( 5 x 18) g of H2O

so, 21.4 g of O2 gives = ( 5 x 18 x 21.4 x 2) / (13 x 32) = 9.25 g of H2O

so, theoretical yields are for CO2 is 18.10 g while for H2O it is 9.25 g