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Topic: Stoichiometry questions  (Read 337 times)

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Offline flowerfluff

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Stoichiometry questions
« on: October 24, 2020, 10:11:44 PM »
I am being a bit confused about how to calculate a mole with a certain decomposition, can someone show me a detailed outlay of how to calculate this so I can learn it?

 Consider the following reaction:

2NaN3     ⟶       2Na +3N2

How many moles of nitrogen gas will be produced by the rapid decomposition of 140g of NaN3?

Offline billnotgatez

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Re: Stoichiometry questions
« Reply #1 on: October 25, 2020, 02:21:01 AM »
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Offline AWK

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Re: Stoichiometry questions
« Reply #2 on: October 25, 2020, 03:47:06 AM »
The Chemistry behind the Air Bag: High Tech in First-Year Chemistry

DOI: 10.1021/ed073p347
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Offline flowerfluff

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Re: Stoichiometry questions
« Reply #3 on: October 25, 2020, 11:17:14 AM »
2x(na)22.99+3x(n)14x2=45.98+84=129.98/2(atomic masses) 64.99

140g/64.99=2.15moles

I tried to redue it by using my book and this is what I gotNaN3 = 1mol/65.02  (na(22.99)+n(14.01)3)=65.01 1mol/2mol

140g x 1/2 molx 42.03 (N)/65.02 = 45.25.....I think this one is closer but Im still unsure how to fix this

« Last Edit: October 25, 2020, 11:50:31 AM by flowerfluff »

Offline AWK

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Re: Stoichiometry questions
« Reply #4 on: October 25, 2020, 11:58:22 AM »
2x(na)22.99+3x(n)14x2=45.98+84=129.98/2(atomic masses) 64.99

140g/64.99=2.15moles
Just read your balanced reaction. Every 2 moles of sodium azide form 3 moles of gaseous nitrogen. You have 2.15 moles of azide.
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Offline flowerfluff

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Re: Stoichiometry questions
« Reply #5 on: October 25, 2020, 12:05:51 PM »
I feel like the way I did it was incorrect. Is that the correct way to do it?

Offline AWK

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Re: Stoichiometry questions
« Reply #6 on: October 25, 2020, 12:19:54 PM »
It is very common for stoichiometry problems to be solved in several correct ways. You're on the right track but I don't think you understand what you figured out at all. This is what happens when you want to solve problems without reading textbooks and training on the examples there.
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