A 4.00 gram sample of aluminum metal is added to a solution containing 5.00 g of

Question

A 4.00 gram sample of aluminum metal is added to a solution containing 5.00 g of the 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 unreactive aluminum is consumed by the reaction with excess HCl. The resulting solution was decanted off and deposited after the metal was washed several times with water. The wet Fe was transferred to a clean, dry evaporating dish which weighed 30.7310 g. After carefully evaporating water from the dish and the iron residue was found to weigh 32.9150 grams

sA Calculate the percentage of iron in the iron chloride

b. determine the empirical formula of the iron chloride

c write a balanced equation for the reaction or iron chloride

d assuming the iron cl is the limiting reagen and using your balanced chemical equation, determin the percentage yield in the experiment/

Explanation / Answer

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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