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I would be grateful,if anyone could clear my doubts.

If we observe this reaction with 4 moles of N2 and 3 moles of H2:

N2 +  3H2  ------>2NH3
4....3.....0  [Before reaction]
4-1/3..0.....3  [After reaction]

So 3 moles of NH3 are formed leavind behind 11/3 Moles of N2.In this reaction the H2 act as limiting reagent and is consumed completely.These reactions are simple to solve as we can easily determine the limiting reagent from the two of the reactants.What if more that 2 reactants are given.Observe the following reaction:

2Fe2S3 + 6H20 + 302-------->4Fe[OH]3 + 6s

If we got 1 mole of Fe2S3,2 mole of H2O and 3 moles of O2.How much moles of Fe[OH]3 will be obtained.This much i could solve in this problem:

2FeS3 + 6H2O + 3O2-------->4Fe[OH]3 + 6S
 1....2....3.....0...0   [Before reaction]
 0....0...3-1-3/2..?...?   [After reaction]

I could not determine which is the reagent among Fe2S3 and H2O to be considered to determine the moles of products as both are the limiting reagents as they are consumed completely leaving behind the oxygen.

My text book says we have to consider H2O as limiting reagent.
So if 6 moles of H2O gives 4 moles of Fe[OH]3 then 2 moles of H2O give 1.34 MOLES OF Fe[OH]3.NOW,WHY H2O WAS CONSIDERED AND WHY NOT Fe2S3?

Is my chemistry book correct about this problem.


Secondly i am messed up with this QUESTION COMPLETELY.

2CoF2 + F2------>2CoF3 and followed by,
[CH2]n + 4nCoF3------>[CF2]n + 2nHF + 4nCoF2

The question is that how much weight of F2 will be consumed to PRODUCE 1Kg OF [CF2]n?

I First combined the two reactions AFTER MULTIPLYING THE 1st REACTION WITH 2n.

This was what i get: 2nF2 + [CH2]n------->[CF2]n + 2nHF

Now what to do next???

I am getting the answer any way but i am not finding my way logical.Answer is 1.52 Kg.This question can also be done by the concept of equivalents:

W of F2 consumed/EQUIVALENT WEIGHT OF F2        =       W of [CF2]n formed/EQUIVALENT WEIGHT OF [CF2]n


I hope someone to help me.

[movies]

Well, let's start at the start.  You've got the stoichiometry wrong in the N2 + H2 equation.  If you started with 4 moles of N2 and 3 moles of H2 you would end up with 3 moles of N2, 0 moles of H2, and two moles of NH3 when the reaction was complete, correct?

[movies]

Next, on the Fe2S3 question, your textbook is correct.

Remember that the limiting reagent is the first one that is completely consumed.  Once you run out of your limiting reagent the reaction won't go anymore and none of your other starting materials will be consumed.  The easiest way to figure out the limiting reagent is to plug the numbers in and calculate how many moles of product you would get given the number of moles of each thing you start with.  So for this example, start with Fe2S3:

You know from the balanced equation that for every two moles of Fe2S3 that are consumed, you will produce 4 moles of Fe(OH)3, assuming that the other reagents are not limiting.  So you can calculate from that molar ratio that if you started with 1 mole of Fe2S3, and Fe2S3 is the limiting reagent, you would end up with 2 moles of Fe(OH)3.

Now do a similar analysis for the other two reactants, assuming that they are the limiting reactant (one at a time).  Whichever one give the _lowest_ number of moles of Fe(OH)3 is your limiting reagent.

[movies]

For the last part calculate the mass of the CF2 formula unit.  Then figure out how many formula units of CF2 you would need to get a kg of material.  Then use the ratio of product (CF2) to reactant (F2) and figure out how many moles of reactant you need.

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Ahhhh...i want to say two moles of NH3 in the products.I apologize.And 3 moles of N2 after reaction.

How could i make such a mistake.Anyway my basics are right but i think i was in a little hurry that day.Sorry to be so foolish.

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See this reaction                                              2Fe2S3 + 6H2O + 3O2-------->4Fe[OH]3 + 6S

Now if we got 1,2,3 moles of Fe2S3,H2O and O2 respectively.How much moles of Fe[OH]3 will be formed?

I know the way to solve the equation.That is equation will proceed it this way


2Fe2S3 + 6H2O + 3O2-------->4Fe[OH]3 + 6S
  1-2/3... 0 ...3-1 ...2x4/3 ..2

The limiting reagent is H2O so 1.43 moles of Fe[OH]3 will be formed leaving behind 0.33 moles of Fe2S3 and 2 Moles of O2.

Now i want to know whether this reaction will proceed as Fe2S3 and 02 are still left?

[movies]

The reaction requires water in order to give the products listed, right?  So if there is no more water, the reaction can't proceed further.

[ssssss]

You mean to say,

Fe2S3 + O2------>No reaction

[movies]

Yup.

According to the equation in the problem water is necessary.

This doesn't preclude some other sort of reaction, however.