If 24.2 g of hydrogen react with excess oxygen and 198 g of water are produced what is the percent yield? *
Your answer

The percent yield of the reaction is 90.9%.

Explanation:

Mass of hydrogen gas = 24.2 g

Moles of hydrogen =

According to reaction, 2 moles of hydrogen gas gives 2 moles of water , then 12.1  moles of hydrogen will give:

 water

Mass of 12.1 moles of water

= 12.1 mol × 18 g/mol = 217.8 g

Theoretical yield of water = 217.8 g

Experimental yield of water = 198 g

The percentage yield of reaction:

The percent yield of the reaction is 90.9%.

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