Question

Question #8d1b2

Answer 1

Start by writing the balanced chemical equation for this reaction

`color(red)(3)NO_(2(g)) + H_2O_((l)) -> color(blue)(2)HNO_(3(aq)) + NO_((g))`

Notice that you have a `color(red)(3):color(blue)(2)` mole ratio between nitrogen dioxide and nitric acid. This means that you can use the number of moles of nitrogen dioxide that react to determine how many moles of nitric acid would be produced.

So, use nitrogen dioxide's molar mass to determine how many moles react

`80cancel("g") * ("1 mole "NO_2)/(46.006cancel("g")) = "1.74 moles"` `NO_2`

That many moles of nitrogen dioxide would produce

`1.74cancel("moles"NO_2) * (color(blue)(2)" moles"HNO_3)/(color(red)(3)cancel("moles"NO_2)) = "1.16 moles"` `HNO_3`

This would be equivalent to producing

`1.16cancel("moles"HNO_3) * "63.01 g"/(1cancel("mole"HNO_3)) = "73 g"` `HNO_3`

This will be the theoretical yield of the reaction, i.e. what is produced when all of the moles of nitrogen dioxide react. This value corresponds to a 100% yield.

Now for the second part. You're told that the reaction actually produces 65 g of nitric acid, so the reaction's percent yield will of course be smaller than 100%.

`"% yield" = "actual yield"/"theoretical yield" * 100`

`"% yield" = (65cancel("g"))/(73cancel("g")) * 100 = color(green)("89 %")`

SIDE NOTE Watch out for the number of sig figs you give for your values. I've used two sig figs for the answers, but I should have used 1 sig fig. For example, 80 g has only 1 sig fig, which means that the theoretical yield should be 70 g.

This will change the percent yield to 93%, so always keep an eye out for sig figs.