You discover a chemical reaction that produces 1 moles of valuable propane for every 6 moles of carbon supplied. essentially, this is an exchange rate of 1 moles of propane for every 6 moles of carbon. if you have 1.8 moles of carbon, how many moles of propane can you produce using this reaction?

Here we have to calculate the number of moles of valuable propane can be prepared from 1.8 moles of carbon.

From 1.8 moles of carbon 0.3 moles of propane can be prepared by the reaction.

From 6 moles of carbon (C) 1 moles of valuable propane (C₃H₈) can be prepared.

Thus from 1.8 moles of C we can obtain ×1.8 = 0.3 moles  of the propane can be prepared.

Thus the amount of propane produced in this reaction is determined.

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