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Topic: Mass-to-Mass Stoichiometry (Gr.11)  (Read 4083 times)

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Offline Rach

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Mass-to-Mass Stoichiometry (Gr.11)
« on: April 23, 2011, 06:34:13 PM »
I am working on a series of homework questions and cannot seem to get the solutions correct. These questions all involve Mass to Mass calculations for  reactants.
An example question reads:
"Ammonium sulfate, (NH4)2SO4, is used as a source of nitrogen in some fertilizers. It reacts with sodium hydroxide to produce sodium sulfate, water and ammonia.
(NH4)2SO4(s) + 2NaOH (aq) ->  Na2SO4(aq) + 2NH3(g) + 2H2O

What mass of sodium hydroxide is required to react completely with 15.4 g of (NH4)2SO4?"

The correct answer is listed as "9.32 g"
My solution was about 14 g...
I have used the n=m/M formula to begin.
Any help would be appreciated

Offline Maverick

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Re: Mass-to-Mass Stoichiometry (Gr.11)
« Reply #1 on: April 23, 2011, 06:55:57 PM »
1. Convert mass of given to moles
2. The multiply moles of given * (coefficient of unknown/coefficient of known).  Sorry for bad explanation I'm on my fone. I would given a better explanation
O and then after that convert back to grams if necessary

Offline Denu

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Re: Mass-to-Mass Stoichiometry (Gr.11)
« Reply #2 on: April 24, 2011, 02:52:41 AM »
This seems like a simple enough question, and I'm usually good at these... I don't know what I'm doing wrong this time. Probably making a stupid mistake. My  answer was 18.67g, and if you divide that by 2, you get approx. 9.33g.. which as you said is the correct answer.

My method: Molar ratio of (NH4)2SO4: NaOH = 1: 2
Mr of (NH4)2SO4: 2(14+4x1) + 32 + 16x4 = 132 g/mol
Number of moles of (NH4)2SO4 according to the given mass of (NH4)2SO4: 15.4/132 = 0.1167 mol
As the molar ratio of (NH4)2SO4 to NaOH is 1:2, the number of moles of NaOH are: 2 x 0.1167 = 0.233 mol
Therefore, mass of NaOH required: Mr(of NaOH) x mol(of NaOH), which is: 2(23+16+1) x 0.233 = 18.7 g.... did I do something wrong? Because you said the answer is 9.32g.. which is the half of 18.7g... You were the one confused, but I`m in the same boat as you now. lol any help would be appreciated.

Offline DrCMS

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Re: Mass-to-Mass Stoichiometry (Gr.11)
« Reply #3 on: April 24, 2011, 03:42:37 AM »
This seems like a simple enough question, and I'm usually good at these... I don't know what I'm doing wrong this time. Probably making a stupid mistake.

Yes and yes

As the molar ratio of (NH4)2SO4 to NaOH is 1:2, the number of moles of NaOH are: 2 x 0.1167 = 0.233 mol
Therefore, mass of NaOH required: Mr(of NaOH) x mol(of NaOH), which is: 2(23+16+1) x 0.233 = 18.7 g.... did I do something wrong?
Yes you have multiplied by 2 twice.  The molecular weight of NaOH is 40.  40 x 0.233 = 9.32

Offline Denu

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Re: Mass-to-Mass Stoichiometry (Gr.11)
« Reply #4 on: April 24, 2011, 12:43:20 PM »
Darn, I knew I was making a stupid mistake. Thanks for pointing it out! :) I feel confident in these type of questions again. I hope the person posting this understood the method too.

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