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Thread: Grams of nitric acid produced/amount of excess reagent remaining?

  1. #1
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    Grams of nitric acid produced/amount of excess reagent remaining?

    How many grams of nitric acid ($\displaystyle HNO_3$) will be produced from the reaction of $\displaystyle 8.25 g$ of $\displaystyle NO_2$ and $\displaystyle 2.25 g$ of $\displaystyle H_2O$ in the following unbalanced reaction:

    $\displaystyle NO_2(g)$ + $\displaystyle H_2O(l) $ $\displaystyle \rightarrow $ $\displaystyle HNO_3(l)$+$\displaystyle NO(g)$

    I balanced it:

    $\displaystyle 3NO_2(g)$ + $\displaystyle H_2O(l) $ $\displaystyle \rightarrow $ $\displaystyle 2HNO_3(l)$+$\displaystyle NO(g)$

    Also, how much of the excess reagent is remaining?

    I haven't done chemistry in a while and am personally not so good at it, but my girlfriend wanted to see if I could get some help for her.
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  2. #2
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    The equation is balanced. First you calculate the mole masses of the substances and get
    No2 = 46g/mol H2O = 18 g/mol HNO3 = 63g/mol.
    Then you calculate out how many moles you have of each of the reactants with M = m / n molmass = grams / mole
    and get NO2 = 0.18 mol H20 = 0.125 mol. And now we also see from youre balanced equation that we need 3 times more moles of NO2 per water. this means that H2O is in excess. So we need 0.06 moles of water. and all the NO2. the excess water is 0.065moles which leaves you 1.17g of water. And you also see that you recive twice the moles of HNO3 as you have water, therefor you recive 0.12 moles HNO3. solving for grams you recive 7.56g of HNO3.
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