[Forumz]

We just learned this lesson today, And I need to know If I'm doing this correctly, and if not, where I went wrong.  Thanks in advance for any replys!

[mike]

Well your answer is right. You could probably set it out a bit more neatly. Also might be good idea to underline your answer.

[Forumz]

Thanks for the quick reply, Mike.  I have noted your suggestions in the next problem. I have attempted the second one, also wondering if it is right. Although, the third one seems to be worded differently, and I would greatly appreciate it if someone could clarify for me what they were asking in relation to the previous two, and what to do differently.  You guys are amaznig, thaks again for your time.  

[mike]

Yes this is right.

If I were doing this though I would not calculate both outcomes. I would simply choose the limiting reagent from the number of moles calculated and use that one in the next calculation of how much product is produced.

For example: you have calculated that there are 0.74 moles of Al and 0.42 moles of Cl2. Now with the mole ratio of 2:3 (or 2Al + 3Cl2) you can already see that the Cl2 must be the limiting reagent.

Good luck

[Forumz]

Thanks again Mike for the assistance for number two, you're a lifesaver.  From now on I'll only calculate the mass of the Limiing Reagent.  And for the third one, I'm not sure of what it's asking, and how it's different from 1 and 2.  But I can see that it's not asking for the same thing.  What is is that I ned to do differently?  (Last question, Promice.lol).  Thanks again

[mike]

It is basically the same (usually in chemistry each question is built on from the previous one). In this example you must write your own balanced equation for the reaction, then the rest is pretty much the same. They also want to know how much of the unused (ie not the limiting reagent) is left after the reaction, so you will have to figure out how much of the non-limiting reagent is used up, and how much is left over.

[Forumz]

k, This is the soution to the Final question on the sheet.  I was curious to as if I had calculated the excess Al correctly?  After this one, I will no longer be bugging you lol .  Thanks again

[mike]

No I don't think this is correct.

Should be:

      2Na     +     Cl2     ---->     2NaCl                          [so this bit is right]

number of moles of sodium = m/M = 10/23 = 0.43 moles (of sodium)

number of moles of chlorine = m/M = 20/71 = 0.28 moles (of chlorine)

As you need half the number of moles of chlorine in order to react with sodium (from your reaction equation 2:1) 0.43 moles of Na would be able to react with only 0.215 moles of chlorine (however you have 0.28 moles so Cl2 must be in excess, right?). So Na is the limiting reagent and Cl2 is in excess.

So how much excess Cl2 is there?

Well as I have just said if all 0.43 moles of sodium react (with 0.215 moles of chlorine) then you will have 0.065 moles of chlorine left over:

0.28 moles (that you had originally) - 0.215 moles (the amount that has reacted) = 0.065 moles (the amount of Cl2 left over)

So how many grams of Cl2 is 0.065 moles?

m = n.M = 0.065 x 71 = 4.6 grams

It sounds confusing but it is not really. You just picked the wrong limiting reagent.

[Forumz]

I believe that I have corrected my stupid errors.  Thank you for all your help, Seems as if I  jumped the gun and Used the wrong mole value for the wrong compound.  This seems to be fully corrected.  As of now I understand how to calculate the excess grams after the reaction had been taken place.  Thanks again for all your help, as I know a major portion of my Grade 11 Chem exam will have a large question on limited reagents.  This forum is truly a gift.