Question

If the percent yield for the following reaction is 75.0%, and 45.0 g of NO2 are consumed in the reaction, how many grams of nitric acid, HNO3(aq), are produced?
3 NO2(g) + H2O(l) ? 2 HNO3(aq) + NO(g)

Answer 1

Hello!

Our first step is to convert the 45.0 grams of NO2 to moles using its molar mass (46 grams per mole):

Next, we use the balanced reaction equation to determine how many moles of Nitric Acid (HNO3) we could have *possibly* produced using 0.98 moles of NO2. From the equation we see that we need 3 moles of NO2 to make 2 moles of HNO3:

Next, we covert 0.65 moles of HNO3 to grams using its molar mass (63 grams per mole):

Last, we take 75% of our theoretical yield (41.1 grams HNO3) to get the mass of HNO3 that was actually produced. This is our final answer:

So, we produced 30.82 grams of HNO3.

I hope this helps! Please rate if it does : )