What is the concentration of bromide, in ppm, if 115.91 g mgbr2 is dissolved in 1.31 l water.

- Bromide concentration is 80748 ppm.

Solution:- ppm is milligram per liter. Grams of magnesium bromide are given and we are asked to calculate the ppm concentration of bromide ion, So, let's first calculate the grams of bromide and then divide it by the given volume.

Molar mass of Magnesium bromide =

= 105.78 g Br

Let's convert grams of Br to milligrams:

since, 1 g = 1000 mg

= 105780 mg

concentration of Br =

= 80748 mg per liter

Since, 1 ppm = 1 mg per liter

So, the bromide concentration is 80748 ppm.

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

the answer would be c