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