Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: help9989 on January 21, 2011, 11:28:53 AM
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A weighed sample of iron (Fe) is added to liquid bromine (Br2) and allowed to react completely. The reaction produces a single product, which can be isolated and weighed. The experiment is repeated a number of times with different masses of iron but with the same amount of bromine. Refer to the following graph.
http://tinypic.com/r/2e4kupj/7 (http://tinypic.com/r/2e4kupj/7)
(a) What mass of Br2 is used when the reaction consumes 2.0 grams of Fe; -->
(b) What is the mole ratio of Br2 to Fe in the reaction;
(c) What is the empirical formula for the product; --> FeBr3
(d) Write a balanced chemical equation for the reaction of iron and bromine; --> Fe + Br2 ---> FeBr3 2Fe + 3Br2 ---> 2FeBr3
(e) What is the name of the reaction product; --> iron(III) bromide
(f) Which statement or statements best describe the experiments summarized by this graph?
(i) When 1.00 grams of Fe is added to the Br2, Fe is the limiting reagent;
(ii) When 3.50 grams of Fe is added to the Br2, there is an excess of Br2;
(iii) When 2.50 grams of Fe is added to the Br2, both reactants are used up completely;
(iv) When 2.00 grams of Fe is added to the Br2, 10.0 grams of product is formed. The percent yield must therefore be 20.0%
Thanks!
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You have to show your attempts at solving the question to receive help. This is a forum policy.
Sure-
(a) From the graph 2 g of Fe makes 11g of product, so joins with 9 g of Br2
(b) this mole ration would depend on the answer for question a.
(c) FeBr3 I think.
(d) Fe + Br2 ---> FeBr3 2Fe + 3Br2 ---> 2FeBr3
(e) iron(III) bromide ?
(f) I believe it would be ii because the other options seem to be incorrect.
Thanks again!
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(a) correct
(b) transfer the weights to moles, then show us what you get
(c-e) depends on b
(f) show your reasoning, other than say: "they are incorrect" Why? what made you say that? (and your answer is not the correct one)
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(a) correct
(b) transfer the weights to moles, then show us what you get
(c-e) depends on b
(f) show your reasoning, other than say: "they are incorrect" Why? what made you say that? (and your answer is not the correct one)
(b) Molar mass (molecular weight) of Br2 is 159.8 g/mol and the Molar mass of Fe is 55.8 g/mol
So is this just asking for the ratio of 9 grams of Br2 compared to 2 grams of Fe?
So 9/159.8= .05 and 2/55.8 = .036. So would the ratio be .05 to .035?
(f) well (iv) can be eliminated because 2.0 grams of Fe yields 11 grams, not 10 grams. Now I am left with (i) (ii) and (iii). And you said it isn't (i), so it must be (ii) or
(iii). I am going to take a shot and guess that it is (ii) .. I really don't understand the concept behind that.
(are c, d and e correct?)
Thanks!
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(b) Molar ratio of Br2 to Fe.
You derive that from your equation 2Fe + 3Br2 ---> 2FeBr3
2 atoms of Fe and 3 molecules of Br2
or looking at it another way :
2 mols of Fe and 3 mols of Br2
(f) Would have multiple answers.
Example : (i) is correct because looking at the graph, 1 g of Fe gives you almost 5 g product.
You add more Fe, you get more product. Therefore 1 g of Fe is not enough.
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you are right with the approach of (b):
Br2: 9/159.8 = 0.056 (or 0.6 mol in 1-significant number); don't forget that 1 mole of Br2 gives 2 moles of Br-
Fe: 2/55.8 = 0.036 (or 0.4 in 1-significant)
(c) correct: this will indeed give you Fe0.4Br1.2: which is more properly written as FeBr3
(d) correct
(e) correct
(f) on the contrary to what vmelkon said, there is only one correct answer, which is indeed answer (I)
(II) is incorrect, as when you add 3.5g of Fe, you are in the area of the graph where no more product is formed if you add more Fe; in other words, the Br2 has reacted all and thus is NOT in excess.
(III) is incorrect as you are once again in the area of the graph where (based on the measured points, NOT the line!) adding more Fe gives no more product. This means Fe is in excess and Br2 has reacted away. (the point where Fe and Br2 are used up completely is at 2g of Fe.)
(IV) is incorrect. 11g of product is formed at 2g of Fe added. But let's disregard that as a typo:
If 2g of Fe reacts with bromide, based on the reaction 2 Fe + 3 Br2 --> 2 FeBr3, 0.04 mol of Fe is expected to make 0.04 mol of product, which is 2(stoechiometry) * 0.04 * 135.8 = 11g
in the graph you can see that indeed 11g has formed, which makes the yield: 11g/11g = 100%, NOT 20%