Look at the chemical equation:

2h2s + 3o2 → 2h2o + 2so2

how many atoms of hydrogen (h) are present on both sides of the equation?

a. 2

b. 3

c. 4

d. 6

4

Explanation:

4 atoms on each size since hydrogen exist as a molecule

C. 4

Explanation:

From the left hand(reaction) side of the equation hydrogen exist as a molecule in hydrogen sulfide(2H2S) . what this simply means it contain two atoms . Now multiplying the bold number in 2H2S  with 2 atoms of hydrogen which constitute a molecule .it will give 4 . On the left Side hydrogen atom(2H2) is 4.

On the right(product) side where Hydrogen appear in the water molecules(2H20), multiply 2 by 2 (2H2) you will get 4 also.

so Hydrogen atom is 4 on both side of the equation.

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