When copper (cu) and oxygen (o) combine, two different compounds are formed. the first compound contains 88.8% copper by mass, and the second compound contains 79.9% copper by mass. (note: assume a 100-gram sample of each compound.) (a) show that cu and o combine with each other in conformity with the law of multiple proportions. the small whole number ratio of the mass of cu per gram of o in the first compound to the mass of cu per gram of o in the second compound equals : 1. this ratio indicates that cu and o with each other in conformity with the law of multiple proportions.

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

law of conservation of matter