Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 29.1 g of butane is mixed with 64. g of oxygen. Calculate the minimum mass of butane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

Question

limiting reactant is butane. .19299moles
1.69238 moles of oxygen gas left over

Answer 1

it appears to me you are definitely off on significant digits.

C4H10+6.5 O2 >>> 5H2O + 4CO2

you start with 29.1/58=.5 moles butane
and 2 moles of O2

for the butane of .5 moles, you need 6.5x.5=3.25 moles O2. You do not have that much, so the limiting reactant is O2.

So if all the O2 is expended, you only use 2/6.5 moles butane

amount of butane left=orig-used
= .5-2/6.5=.29 moles butane left over.

So check all that, it is certainly much different than what you got. I did most of the math in my head.

Answer 2

2C4H10 + 13O2---> 8CO2 + 10H2O

Rmm of 1 mole oxygen=32g

13 mole=416g

mass of 1 mole of butane=58g
2 mole=116g

116g C4H10 requires 416g O2

17.8 C4H10 will require 64g O2

excess butane=29.1-17.8=11.3g