[star4]

I'm currently a high school chemistry ap student and our teacher has given us an extra credit problems that are very hard. If anybody could help me on these, it would be a great help.

1. Iron filings were combined with excess chlorine gas to produce Iron(III) chloride, however, a small quantity of Iron(II) was also produced. A masss of 2.55 grams of iron reacted to form a total of 6.99 grams of mixed iron chlorine product (ie, Iron (II) chloride and Iron(II) chloride). What was the percent yield of Iron (III) chloride from this process?

2. Water is added to 4.267 grams of UF6. The only products are 3.730 grams of a solid containing only uranium, oxygen and fluorine and .970 gram of a gas. The gas is 95% fluorine and the remainder is hydrogen.

(a) From this  data, determine the empirical formula of the gas.
 i got that part, HF.
(b) what fraction of the fluorine of the original compound is in the solid and what fraction in the gas after the reaction?
(c) What is the formula of the solid product?
(d) Write a balanced equation for the reaction between UF6 and H20. Assume that the empirical formula of the gas is the true formula.

[Padfoot]

1. Iron filings were combined with excess chlorine gas to produce Iron(III) chloride, however, a small quantity of Iron(II) was also produced. A masss of 2.55 grams of iron reacted to form a total of 6.99 grams of mixed iron chlorine product (ie, Iron (II) chloride and Iron(II) chloride). What was the percent yield of Iron (III) chloride from this process?

Hint: try using simultaneous equations.  Do you know what these are?
http://en.wikipedia.org/wiki/Simultaneous_equations

2. Water is added to 4.267 grams of UF6. The only products are 3.730 grams of a solid containing only uranium, oxygen and fluorine and .970 gram of a gas. The gas is 95% fluorine and the remainder is hydrogen.

(a) From this  data, determine the empirical formula of the gas.
 i got that part, HF.
(b) what fraction of the fluorine of the original compound is in the solid and what fraction in the gas after the reaction?
(c) What is the formula of the solid product?
(d) Write a balanced equation for the reaction between UF6 and H20. Assume that the empirical formula of the gas is the true formula.

The reaction is:
UF6 + H20 → UO2F2 + HF   (not balanced though)
Using the reaction eqn, the problem is like any other stoichiometry problems.

I'll let you earn your extra credit by balancing it yourself

[star4]

ummm how did u find that reaction?

[Padfoot]

http://cameochemicals.noaa.gov/chemical/14922

[Padfoot]

Wait! I think we cheated, we're probably not allowed to just hunt down the reaction eqn.

Ok, I think I know how to do it, I'll get you started:

Calculate quantity of O in solid product:
0.97g HF= 0.0485mol HF formed
0.0485/2= 0.02425mol H20 reacts  (Since other product has no H)
From above, 0.02425mol of O in solid product (only product with O)

From the amount of UF6 reacted, you can work out how much U is in the solid product since other product has no U.
n(UF6)=4.267/352=0.01212mol

Given that you have 3.73g of solid product you can now solve for the quantity of F in the solid (you could also use the amount of HF formed and the amount of UF6 reacted).

b), c) and d) can be solved from here.

Hope that Helps