Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: flowerfluff on October 24, 2020, 10:11:44 PM
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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?
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The Chemistry behind the Air Bag: High Tech in First-Year Chemistry
DOI: 10.1021/ed073p347
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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
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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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I feel like the way I did it was incorrect. Is that the correct way to do it?
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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.