What is the oxidation number for chromium in each of these compounds. show your work. a. k2cro4 b. cr2o3

A. + 6.

B. + 3.

Explanation:

A. K₂CrO₄:

We have that the sum of the oxidation no. of different elements in the compound multiplied by its no. is equal to the overall charge of the compound.

∵ 2(oxidation no. of K) + (oxidation no. of Cr) + 4(oxidation no. of O) = 0.

oxidation no. of K = + 1, oxidation no. of O = - 2.

∴ 2(+ 1) + (oxidation no. of Cr) +4(- 2) = 0.

∴ 2 + (oxidation no. of Cr) - 8 = 0.

∴ (oxidation no. of Cr) - 6 = 0.

∴ oxidation no. of Cr = + 6.

B. Cr₂O₃:

∵ 2(oxidation no. of Cr) + 3(oxidation no. of O) = 0.

oxidation no. of O = - 2.

∴ 2(oxidation no. of Cr) + 3(- 2) = 0.

∴ 2(oxidation no. of Cr) - 6 = 0.

∴ 2(oxidation no. of Cr) = + 6.

∴ (oxidation no. of Cr) = + 6/2 = + 3.

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

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