Madchemyst
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Stoichiometry Question
I want to make some AgNO3. I have 3.31 g of Silver and I would like to use the stoichiometry amount of HNO3 for the reaction. Here is what I think
is the correct answer but it does not seem right. I am new to all this so if I am way off I would like to understand the correct way. Thanks much
3.31 g of Ag = .03 moles
I will be using 70 % HNO3
The Molar Mass of HNO3 = 63.01 g/mol
.03 g of HNO3 = 1.89 g
Moles/Liter of HNO3 = 1413 g/l
70% HNO3 = 1413 * .7 = 989.1 g HNO3/L
1.89g (1L/989.1g) * 1000ml/1L = 1.9 ml needed of 70% HNO3.
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ISCGora
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Umm maybe isnt correct I didnt in hurry I got 2.509g of 70% HNO3.
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Milan
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ISCGora you were close it's 2.762 g of 70% HNO3.
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gdflp
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Unfortunately, no this is incorrect. Silver is lower on the reactivity series than hydrogen, thus the following reaction does not occur : Ag +
HNO3 --> AgNO3 + 1/2H2.
Theoretically you could achieve: 3Ag + 4HNO3 --> 3AgNO3 + 2H2O + NO but the rate of the reaction will be
ridiculously slow as the nitric acid needs to be kept cold to achieve this stoichiometry. To get a reasonable reaction rate, you will need to heat up
the nitric acid and will get the following stoichiometry Ag + 2HNO3 --> AgNO3 + H2O + NO2. Thus you
need a molar ratio of Ag:HNO3 1:2.
Then, assuming a density of 1413g/L * 0.7 = 989.1g HNO3/L = 15.7mols/L. Now you need .06mols of nitric acid so .06/15.7 = 0.0038L = 3.8mL
of 70% nitric acid.
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Milan
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Then if the ratio is 1:2 then you would need 5.525 g of 70% HNO3.
Edit: I just did the calculation. If the temperature of the 70% HNO3 is 20°C then it's density is 1.41340 g/mL which gives us 7.808 mL of 70% HNO3.
[Edited on 6-4-2015 by Milan]
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gdflp
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Quote: Originally posted by Milan  | Then if the ratio is 1:2 then you would need 5.525 g of 70% HNO3.
Edit: I just did the calculation. If the temperature of the 70% HNO3 is 20°C then it's density is 1.41340 g/mL which gives us 7.808 mL of 70% HNO3.
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You need to divide by density, not multiply giving 3.9mL. I rounded slightly, but if you're actually going to run the reaction, at least a 10% excess
should be used to account for evaporating nitric acid and to keep the concentration reasonably high throughout the reaction.
[Edited on 4-6-2015 by gdflp]
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Milan
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Wow, can't believe I misfired the button , maybe I should use the keyboard next
time.
Well anyway, I did it again. You were right it's 3.9 mL, that is in room temperature. Sorry for the mistake.
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Madchemyst
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gdflp, thanks for the explanation. My first mistake was not showing the products from the reaction and although I would have missed the 1/2H2. It now
makes sense to me since you showed it. But how would you know that heating the HNO3 would cause the reaction to go? I see that in order for the
reaction to work 2 molecules of HNO3 would be needed since Hydrogen is diatomic and that would then mean that H2O and NO would also be produced.
Lots to learn! I also understand the NO2 is not good stuff so I need to do this in a fume hood or outdoors.
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