How many moles of co2 form when 58.0 g of butane, c4h10, burn in oxygen? 2c4h10 + 13o2 → 8co2 + 10h2o a. 4 mol b. 2 mol c. 8 mol d. 10 mol

The molar mass of butane is 12 * 4 + 1 * 10 = 58 grams, so we have one mole of butane.  Because this reaction creates eight molecules of CO2 for every two molecules of butane, it creates four CO2 molecules for every butane molecule, so, because we only have one mole of butane, it creates four moles of CO2.

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g