How many miles of fe2+ ions and mno4- ions were titrated in each part 1 trial

In 1000 ml there is 0.10 moles of Fe 2+

Therefore, in 10 ml there is (0.1/1000)*10= 0.001 mol of Fe2+

mole ratio for rxn Fe2+ : MnO4- is

1 : 2

therefore if 0.001 moles of Fe2+ react then 0.001*2 =0.002 moles of MnO4- react with Fe2+

hence, molarity of MnO4- = (mol*vol)/1000

= 0.002*10.75/1000= 2.15*10-5M

Explanation:

Hope this helps

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

i believe the answer is three water molecules