Chemical Forums

Chemistry Forums for Students => High School Chemistry Forum => Topic started by: MakoEyes on September 26, 2009, 06:38:22 PM

Title: Finding how many moles needed to complete reaction?
Post by: MakoEyes on September 26, 2009, 06:38:22 PM
Here's a question I don't know the technique I would use to solve it. Can someone offer me some solution so I can observe the techniques used?

The elements nitrogen and magnesium react to give magnesium nitride according to the balanced
equation:
N2 + 3 Mg ->  Mg3N2
Suppose you have 0.4530 mol N2.
(a) How many moles of Mg are needed for complete reaction?
(b) What mass of Mg3N2, in grams, can be produced? (The molar mass of Mg3N2 is 100.9284 g/mol).
Title: Re: Finding how many moles needed to complete reaction?
Post by: Borek on September 26, 2009, 07:21:00 PM
Do you know how to read a reaction equation?

http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations
Title: Re: Finding how many moles needed to complete reaction?
Post by: MakoEyes on September 26, 2009, 07:38:18 PM
Yes I do.

For (a), is this the technique:

0.4530 mol N2 * (3 mole Mg / 1 mole N2) = answer

? Or am I off base a bit.
Title: Re: Finding how many moles needed to complete reaction?
Post by: Ak on September 26, 2009, 11:14:04 PM
yeah thts right and then for (b) you take either the moles of N2 or Mg (id go with N2 cuz its 1:1).  Then u have g/mol and mol and u need to get grams (g).
Title: Re: Finding how many moles needed to complete reaction?
Post by: MakoEyes on September 27, 2009, 10:50:07 PM
Quote from: Ak on September 26, 2009, 11:14:04 PM
yeah thts right and then for (b) you take either the moles of N2 or Mg (id go with N2 cuz its 1:1).  Then u have g/mol and mol and u need to get grams (g).
I'm a bit confused by your reply.

Here is what I did for (b)...

0.4530 mole N2 * (14.01g / 1 mole N) * (1 mole N2 / 1 mole Mg3N2) * (100.9284g / 1 mole Mg3N2) = 640.5g Mg3N2

Is this correct or did I do this wrong?
Title: Re: Finding how many moles needed to complete reaction?
Post by: Borek on September 28, 2009, 03:36:21 AM
It is wrong. But dimensional analysis is not my forte.
Title: Re: Finding how many moles needed to complete reaction?
Post by: cliverlong on September 28, 2009, 04:14:22 AM
Quote from: MakoEyes on September 27, 2009, 10:50:07 PM

Here is what I did for (b)...

0.4530 mole N2 * (14.01g / 1 mole N) * (1 mole N2 / 1 mole Mg3N2) * (100.9284g / 1 mole Mg3N2) = 640.5g Mg3N2

Is this correct or did I do this wrong?
Your lack of use of subscripts hurts my eyes - but that may be of no concern to you. More importantly, that, the arrangement of your units and not splitting the calculation into steps, makes it difficult to see where the problem is in your answer.

Rearranging your information I think reveals where the problem may lie. Your calculation:

0.4530 N2 [mole] * (14.01 /1 [g / mole] ) * (1 N2 / 1  Mg3N2 [mole / mole]) * (100.9284/ 1  Mg3N2 [g/mole]) = 640.5g Mg3N2

Looking at your left-hand side, the units work out as

[mole] * [g / mole]  * [mole / mole] *  [g/mole] =  [g2/mole]. There is a problem there.

I would look back at the single, long string of multiplications and split them into a series of steps, checking the units as I went

Clive
Title: Re: Finding how many moles needed to complete reaction?
Post by: MakoEyes on September 30, 2009, 11:59:15 PM
Thanks Clive.. I still can't figure it out tho. I can't seem to "develop" the dimensional analysis needed to solve this problem.

Could someone show me the steps? (for b)
Title: Re: Finding how many moles needed to complete reaction?
Post by: cliverlong on October 01, 2009, 02:08:53 AM
Quote from: MakoEyes on September 26, 2009, 06:38:22 PM
N2 + 3 Mg ->  Mg3N2
Suppose you have 0.4530 mol N2.
(a) How many moles of Mg are needed for complete reaction?
Well,

From the equation you can see every 1 mole of N2 will react with 3 moles of Mg.

So how many moles of Mg will react with 0.4530 mol N2?
Quote
(b) What mass of Mg3N2, in grams, can be produced? (The molar mass of Mg3N2 is 100.9284 g/mol).
Moles again.

From the equation you can see every 1 mole of N2 will produce 1 mole of  Mg3N2

So how many moles of  Mg3N2 will be prdoced from 0.4530 mol N2?

What is the equation that will convert moles into mass? (you are given molar mass of Mg3N2)
Title: Re: Finding how many moles needed to complete reaction?
Post by: MakoEyes on October 01, 2009, 02:39:05 PM
Thus:


0.4530 mol N2 * (3 Mole Mg / 1 Mole N2) * (100.9284 g/mol Mg2N3 / 1 mole Mg2N3) = answer?