Industrially, nitric acid is produced by the Ostwald process...

Industrially, nitric acid is produced by the Ostwald process represented by the following equations: $$\begin{array}{c} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HNO}_{3}(a q)+\mathrm{HNO}_{2}(a q) \end{array}$$ What mass of $\mathrm{NH}_{3}$ (in grams) must be used to produce 1.00 ton of $\mathrm{HNO}_{3}$ by the Ostwald process, assuming an 80 percent yield in each step (1 ton $=2000$ lb; 1 b $=453.6$ g)?

Industrially, nitric acid is produced by the Ostwald process represented by the following equations:
$$\begin{array}{c}
4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\
2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \\
2 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HNO}_{3}(a q)+\mathrm{HNO}_{2}(a q)
\end{array}$$
What mass of $\mathrm{NH}_{3}$ (in grams) must be used to produce 1.00 ton of $\mathrm{HNO}_{3}$ by the Ostwald process, assuming an 80 percent yield in each step (1 ton $=2000$ lb; 1 b $=453.6$ g)?

Key Concepts

Percent Yield

Percent yield is the efficiency metric used in chemical reactions that compares the actual amount of product obtained to the theoretical maximum possible based on stoichiometric calculations. It accounts for losses and inefficiencies in each reaction step, which is critical in industrial production where multiple steps and yields compound.

Ostwald Process

The Ostwald process is an industrial chemical procedure for producing nitric acid via the catalytic oxidation of ammonia. It involves the sequential conversion of ammonia to nitric oxide, then to nitrogen dioxide, and finally its reaction with water to form nitric acid. This process exemplifies how intermediates in reaction sequences are managed to obtain the desired final product.

Stoichiometry

Stoichiometry is the quantitative study of reactants and products in chemical reactions. It involves using balanced chemical equations to determine the relative amounts of substances consumed and produced, allowing for calculations of mass and moles. This concept is essential in designing and scaling up industrial chemical processes.

Molar Mass and Unit Conversions

Molar mass and unit conversions are fundamental techniques used to link the molar scale of chemical reactions to measurable quantities like grams or pounds. This concept involves converting between mass units and moles, ensuring that calculations accurately relate quantitative laboratory or industrial measurements to theoretical stoichiometric predictions.

Transcript

00:01 So for this problem, we have a process described as the oswald process, and we want to find the mass of ammonia required to produce a ton of nitric acid.

00:10 The key here is that there's an 80 % yield in each step.

00:13 So we have to work backwards three times to find our ammonia.

00:18 So the first thing that we want to do is you want to find the theoretical yield of nitric acid, given that we know its actual yield.

00:26 So we know that the actual yield is going to be one ton.

00:35 And we know that the formula is going to be the actual yield over the theoretical yield, which is what we're trying to find, times 100.

00:46 And we know that that yield is going to be 80 % as given to us in the problem.

00:51 So all i'm going to do is i'm going to rearrange this equation.

00:54 So i know that x is equal to 1 ton times 100 % over 80%.

01:04 And when i multiply that through, i get the 1 .25 tons.

01:10 Is my theoretical yield for nitric acid.

01:15 So now that i have my theoretical yield, i'm going to have to do some stoichiometry to then calculate the grams of no2, which is a reactant for this process.

01:25 So i'm going to have to find the molar masses of both of these substances.

01:29 So first we have nitric acid.

01:34 Well, i have one hydrogen, which is 1 .008.

01:37 We have a nitrogen, which is 14 .01.

01:40 And i have three oxygens which are each 16 .1.

01:49 And when i multiply all of these through, i get that my molar mass is just going to end up being 63 .012.

02:06 Now i want to do n02.

02:09 For n02, i have one nitrogen, which is 14 .01.

02:13 And i have oxygen, which is 16 .00.

02:16 But i have two of them, so i multiply this by two.

02:20 And i multiply it through, and i get that my molar mass is end up going to be 46 .01 grams per mole.

02:30 So these are to my two molar masses that i have, and now i can do stoichiometry to go back.

02:36 So i know that i need to produce 1 .25 tons of hn .03.

02:46 Well, i know that there's 2 ,000 pounds in one ton, and as given to us in the problem, we know that there are 453 .6 grams in every pound.

03:00 Now, you just divide by the molar mass of hno3 that we calculated, which is 63 .012 grams per mole.

03:12 And now i multiply by the multiple ratio.

03:16 Well, if i look in my initial equation that i'm given in the problem, i see that two moles of no2 are required to produce 1 mole of hno3 so i know that my mole to mole ratio is just going to end up being 2 to 1.

03:29 So i can continue this here.

03:32 I know 2 mole of n02 is required for 1 mole of hno3.

03:40 And then finally i just multiply by the molar mass of n02 which we calculated to be 46 .01 grams per mole.

03:51 And i will get 1 .66 times 10 to the 6 grams of no2 is required.

04:01 But now once again i have to find the theoretical yield because this represent right here represents the actual yield.

04:08 Now i have to take into account that only 80 % of that is going to be netted.

04:14 So now i go back to the second equation.

04:17 So, and we're going to use the exact same formula like we did before.

04:20 This represents the actual yield.

04:23 So i have 1 .66 times 10 to the 6 grams...