Zenneth M.

# When 70mL of a 2.5M nitric acid solution is reacted with 11g of copper metal, what is the most amount of nitrogen dioxide gas that is able to be produced?

4HNO3+Cu(s) --> 2H2O + Cu(NO3)2 + 2NO2

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Zenneth M.

So are there .173 moles of Cu and .692 moles of nitric acid in the equation?
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12/04/20

Greg S.

Your math on Cu is correct; and, to state your comment more accurately, 0.692 moles of nitric acid are *required* for a 4:1 molar ratio. So, you should ask yourself: is that much nitric acid provided/available? If there is more than that, then Cu is the limiting reagent, as you would run out of that first. If there is less than that, then you would run out of nitric acid first. Once you know the limiting reagent, you can then calculate how much product is formed.
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12/04/20

Zenneth M.

Oh, so in this case Cu is the limiting reagent, right? Since 70mL of a 2.5M solution of nitric acid is 0.175 moles, with the ratio it comes out to 0.7 moles, which is greater than 0.692 moles.
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12/04/20

Greg S.

Good try, but actually, you have that reversed. 0.173 moles of copper will *require* 0.692 moles of nitric acid. Because you only have 0.175 moles of nitric acid (which you calculated correctly), it is the limiting reagent. It will be enough to react with only 0.175/4 = 0.438 moles of copper.
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12/04/20

Zenneth M.

Oh, right, that makes more sense now. Thank you for your help.
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12/04/20

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