chemistry

in an experiment, a 4.000grams sample of aluminum metal is added to a solution containing 5.000grams of dissolved iron chloride. the reaction results in the formation of aluminum chloride and metallic iron. when the reaction is complete, unreacted aluminum remains, and this unreacted aluminum is consumed by reaction with excess hydrochloric acid. the resulting solution was decanted off and the deposited Fe metal was washed several times with water. the wet Fe was transferred to a clean, dry evaporating dish, which weighted 30.7310grams. after carefully evaporating the water, the dish and the iron residue was found to weigh 32.9150grams.

A. calculate the experimental percentage of iron in the iron chloride

B. determine the empirical formula of the iron chloride

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1. Second try. My first one was deleted.
32.9150g = mass dish + Fe
-30.7310g = mass empty dish
-------------
+2.1840g = mass Fe
%Fe = (mass Fe/mass Fe sample) = 2.1840/5.000)*100 = ? = approx 43.68%

Note: I feel compelled to point out that this is a lousy procedure BECAUSE when the HCl is added to consume the unused Al metal, it will consume part of the solid Fe metal ALSO which means the results are low for the percentage of Fe.

b) Take a 100 g sample which give you
43.68 g Fe and
100 - 43.68 = 56.32g Cl

mols Fe = 43.68/55.85 = about 0.782
mols Cl = 56.32/35.45 = about 1.59
Divide both by the smaller of the two or
0.782/0.782 = 1.00 and
1.59/0.782 = 2.03 which I would round to 2 which makes the iron chloride empirical formula FeCl2.

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2. what is the balanced chemical equation and how do you find the percent yield of iron in the experiment?

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