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Topic: How to calculate the perfect ratios for reactions?  (Read 821 times)

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Offline Richeyyy

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How to calculate the perfect ratios for reactions?
« on: September 22, 2021, 02:38:20 PM »
Hello, I wanna know how to calculate stoichiometric ratios for reactions. I know that stoichiometric ratios are basically balanced out reactions converted into percentages. I know how to balance out chemical equations, and I know that there has to be an equal amount of moles for each reactant for a stoichiometric ratio. How do I convert "that" into percentage though? I can't seem to find any proper tutorials on that

Offline Babcock_Hall

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Re: How to calculate the perfect ratios for reactions?
« Reply #1 on: September 22, 2021, 02:51:33 PM »
I do not understand your question.  Can you clarify what you mean by percentages?

Offline Richeyyy

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Re: How to calculate the perfect ratios for reactions?
« Reply #2 on: September 22, 2021, 10:02:58 PM »
I do not understand your question.  Can you clarify what you mean by percentages?
Yeah I kinda did make it confusing back there, sorry. It's just that I wanna learn how to calculate the stoichiometric ratio for kno3 + sugar. The equation is 2 KNO3(s) + CH2O(s) → 2 KNO2(s) + CO2(g) + H2O(g). Knowing this, I wanna know how much of kno3 and sugar to add in percentage. And I dunno how to do that. Example 65% kno3 and 35% sugar.

Offline Borek

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Re: How to calculate the perfect ratios for reactions?
« Reply #3 on: September 23, 2021, 02:52:10 AM »
if you know molar ratio converting it to mass ratio (hint: molar mass) is trivial. Mass ratio can be always expressed as percentage (hint: if you mix 10g of this with 30g of that, mass percentage of this is 10/(10+30)×100%, just apply the definition).

Note: sugar/nitrate mixture borders pyrotechnics, we don't discuss pyrotechnics at the forum.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline Richeyyy

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Re: How to calculate the perfect ratios for reactions?
« Reply #4 on: September 23, 2021, 03:06:07 AM »
Oh? Thanks so much for the info, idk I don't view it as something as serious as pyrotechnics, just a basic experiment for fun

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