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Corn farmers in the American Midwest typically use $5.0 \times 10^{3}$ kilograms of ammonium nitrate fertilizer per square kilometer of cornfield per year. Some of the fertilizer washes into the Mississippi River and eventually flows into the Gulf of Mexico, promoting the growth of algae and endangering other aquatic life. a. Ammonium nitrate can be prepared by the following reaction: $$\mathrm{NH}_{3}(g)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{NH}_{4} \mathrm{NO}_{3}(a q)$$ How much nitric acid would be required to make the fertilizer needed for $1 \mathrm{km}^{2}$ of cornfield per year? b. Ammonium ions dissolved in groundwater may be converted into $\mathrm{NO}_{3}^{-}$ ions by bacterial action: $$\mathrm{NH}_{4}^{+}(a q)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{NO}_{3}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)+2 \mathrm{H}^{+}(a q)$$ If $10 \%$ of the ammonium component of $5.0 \times 10^{3}$ kilograms of fertilizer ends up as nitrate ions, how much oxygen would be consumed?

          Corn farmers in the American Midwest typically use $5.0 \times 10^{3}$ kilograms of ammonium nitrate fertilizer per square kilometer of cornfield per year. Some of the fertilizer washes into the Mississippi River and eventually flows into the Gulf of Mexico, promoting the growth of algae and endangering other aquatic life.
a. Ammonium nitrate can be prepared by the following reaction:
$$\mathrm{NH}_{3}(g)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{NH}_{4} \mathrm{NO}_{3}(a q)$$
How much nitric acid would be required to make the fertilizer needed for $1 \mathrm{km}^{2}$ of cornfield per year?
b. Ammonium ions dissolved in groundwater may be converted into $\mathrm{NO}_{3}^{-}$ ions by bacterial action:
$$\mathrm{NH}_{4}^{+}(a q)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{NO}_{3}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)+2 \mathrm{H}^{+}(a q)$$
If $10 \%$ of the ammonium component of $5.0 \times 10^{3}$ kilograms of fertilizer ends up as nitrate ions, how much oxygen would be consumed?

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Added by Michael D.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

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Solved by Expert Susan Hallstrom

01/28/2023

Step 1

Step 1: The ammonium nitrate can be prepared by the following reaction: $$\mathrm{NH}_{3}(g)+\mathrm{HNO}_{3}(a q) ightarrow \mathrm{NH}_{4} \mathrm{NO}_{3}(a q)$$ In this reaction, $\mathrm{NH}_{4} \mathrm{NO}_{3}(a q)$ is the nitrate ion. Show more…

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Corn farmers in the American Midwest typically use $5.0 \times 10^{3}$ kilograms of ammonium nitrate fertilizer per square kilometer of cornfield per year. Some of the fertilizer washes into the Mississippi River and eventually flows into the Gulf of Mexico, promoting the growth of algae and endangering other aquatic life. a. Ammonium nitrate can be prepared by the following reaction: $$\mathrm{NH}_{3}(g)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{NH}_{4} \mathrm{NO}_{3}(a q)$$ How much nitric acid would be required to make the fertilizer needed for $1 \mathrm{km}^{2}$ of cornfield per year? b. Ammonium ions dissolved in groundwater may be converted into $\mathrm{NO}_{3}^{-}$ ions by bacterial action: $$\mathrm{NH}_{4}^{+}(a q)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{NO}_{3}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)+2 \mathrm{H}^{+}(a q)$$ If $10 \%$ of the ammonium component of $5.0 \times 10^{3}$ kilograms of fertilizer ends up as nitrate ions, how much oxygen would be consumed?

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Transcript

00:08 Farmers typically used 5 .0 times 10 to the third kilograms of ammonium nitrate per one kilometer squared per year.

00:33 I don't think i need that.

00:34 And then i've got this chemical equation right here and we're asked how much nitric acid would be needed to make this.

00:48 Okay.

00:48 Per year.

00:53 Okay.

00:54 So i've got 5 .0 times 10 to the third for one km squared.

01:05 10 to the third kilograms in one km squared.

01:14 And we're figuring out one km squared.

01:16 So i'm just going to put that up there.

01:20 And i know that i would need that many kilograms of, i should have been clear here.

01:31 That many kilograms of ammonium nitrate.

01:43 Okay.

01:45 And how much nitric acid? i'm going to probably report this in kilograms.

01:56 Since i'm in kilograms and i'm going two kilograms, i can leave it in kilograms.

02:01 So my molar mass of ammonium nitrate will be 80 .04 grams per mole of ammonium nitrate.

02:30 And all of my ratios were one to one.

02:35 And i'll have nitric acid to ammonium nitrate.

02:41 And finally, my molar mass of nitric acid will be not very much.

02:52 48, 48, 68, 63...

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