To determine the percentage of bromide in a mixture containing a soluble bromide salt, the procedure from Experiment 10 is used except the bromide is precipitated with aqueous lead (II) nitrate, Pb(NO3)2, forming PbBr2(s). Initially, a 0.5014 g sample of an unknown bromide salt is dissolved in water. Excess Pb(NO3)2 is added to the solution to quantitatively precipitate the bromide from the solution. The resulting mixture is filtered through a piece of filter paper weighing 1.3258 g. After several days of drying, the filter paper and precipitate weighed 1.7895 g. Calculate the percent bromide in the original sample.

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 is minerals.

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa