If 40.0g of nitric acid is produced by reacting 100g of NO2(g), what is the percent yield of of this reaction?

Answer 1

The relative molar mass of NO2 is Mr = 46 so the 100 g of NO2 are:

n = m/Mr, n = 100/46 mol = 2.174 mol (approx.)

According to the chem. equation

3NO2 + H2O --> 2HNO3 + NO

3 mol of NO2 give 2 mol of HNO3, so
2.174 mol mol of NO2 give: 2*2.174/3 = 1.45 mol of HNO3

The relative molar mass of HNO3 is Mr = 63, so theoretically you get:

m = n*Mr = 1.45*63 = 91.3 g of HNO3

But, you got 40 g of HNO3, so the yield is:

100*40/91.3 = 43.81%

Answer 2

For the extremely steps in contact convert the grams of each and every factor into moles applying molecular weights. applying the mole ratios from the balanced equation, discover the proscribing reagent, and in line with that decide the the % yield.

Answer 3

percent yield is the theoretical yield minus the actual yield over the theoretical yield multiplied by 100. therefore your question is impossible to answer due to the fact that you haven't done an experiment.

Answer 4

reacting 100g of NO2(g) with what?

Nitric acid is HNO3. That's about all I can tell you.

Answer 5

40 percent