HELP! stoichiometry has gotten the best of me

Question

Nitrogen dioxide gas and water are placed in a closed container and react to form an aqueous solution

of nitric acid and nitrogen monoxide gas under STP conditions. The following data table represents

lab data acquired from this reaction.

Initial amount of

nitrogen dioxide in the

container: 112 L

Initial amount of water

in the container 40.0 mL (40.0 g)

Amount of nitric acid

produced 142 g

a. Write the balanced equation for this reaction including the states of matter.

b. Assuming the reaction goes to completion, calculate the theoretical yield, in grams, of nitric

acid according to the conditions provided above.

c. Calculate the percent yield of the experiment.

d. Based on your calculation in part c, suppose a company is looking to produce 25.0 kg of nitric

acid according to the same reaction and experimental setup. Assuming there is excess water

in the container available, calculate the minimum liters of nitrogen dioxide needed.

e. Calculate each of the following:

Amount of excess reagent, in grams, remaining:

Amount of nitrogen monoxide produced in grams:

f. Calculate the density, in g/L, of the nitrogen dioxide gas at STP.

g. Suppose a reaction is performed to remove the hydrogen from nitric acid. How many grams

of hydrogen could be obtained from the nitric acid produced as a result of the experiment?

Answer 1

Hello, Jonathon,

a. Write the balanced equation for this reaction including the states of matter.

b. Assuming the reaction goes to completion, calculate the theoretical yield, in grams, of nitric acid according to the conditions provided above.

The NO₂ gas is at STP, so we can use the conversion factor 22.4L/mole.

NO₂ = 112L/(22.4L/mole) = 5 moles NO₂

The equation tells us we should get 2 moles HNO₃ per 3 moles NO₂. So we are expecting (5 moles NO₂)*(2 Moles HNO₃/3 Moles NO₂) or 3.333 moles HNO_3. Multiply by the molar mass of HNO₃ (63 g/mole) to rfind grams HNO₃. I get 210 grams HNO₃. We actually got 142 grams.

c. Calculate the percent yield of the experiment.

That is a (142g/210g) or 67.6% yield.

d. Based on your calculation in part c, suppose a company is looking to produce 25.0 kg of nitric

acid according to the same reaction and experimental setup. Assuming there is excess water

in the container available, calculate the minimum liters of nitrogen dioxide needed.

25.0 kg of HNO₃ is (25,000g)/(63 g/mole) = 397 moles of HNO₃. It takes 3 moles of NO₂ to make 2 moles of HNO₃, so we need to multiply the moles of HNO₃ by 3/2 to find the moles of NO₂ theorectically needed (595 moles of NO₂). BUT, we only got a 67.6% yield. So we need to increase the moles of NO₂ by that amount ((595 moles of NO₂)/0.676) = 880 moles.

e. Calculate each of the following:

Amount of excess reagent, in grams, remaining:

We have more than enough H₂O for the reaction. I find an excess of 0.556 moles after comparing the moles needed for reacting with 5 moles of NO₂. That converts to 10 grams of water.

O consumed.

Amount of nitrogen monoxide produced in grams:

We should get 50 grams theorectically because the moles of NO will equal the moles H₂O consumed. But the actual yield is 67.7%, so it needs to be reduced by that amount. I get 50 grams, but please check your calculations - I'm getting lost in all these numbers.

f. Calculate the density, in g/L, of the nitrogen dioxide gas at STP.

1 mole is 22.4 liters. NO₂ has a molar mass of 46 g/mole. So the density is 46 g/22.4 liters. That is 2.05 g/L.

g. Suppose a reaction is performed to remove the hydrogen from nitric acid. How many grams of hydrogen could be obtained from the nitric acid produced as a result of the experiment?

We produced 142 grams of nitric acid, HNO₃. The mass percentage of hydrogen in the molecule is 1/63, or 0.01587. The mass of H produced would be (0.01587)*(142 grams),or 2.25 grams.

There are a lot of calculations, so check my work.

I hope this helps,

Bob