The fertilizer, ammonium nitrate, is made by reacting...

The fertilizer, ammonium nitrate, is made by reacting ammonia with nitric acid. (Section $1.4)$ (a) Write a balanced equation, with state symbols, for the reaction of ammonia gas with nitric acid to form a solution of ammonium nitrate. (b) Rewrite the equation to show the ions present in the reactants and products, and hence write an ionic equation for the reaction. (c) What type of chemical reaction does your equation represent? Nitric acid is manufactured from ammonia. The first stage in this process involves burning ammonia in oxygen on the surface of a platinum gauze catalyst. The products are $\mathrm{NO}$ and $\mathrm{H}_{2} \mathrm{O}$. (d) Construct a balanced equation for the buming of $\mathrm{NH}_{3}$ in $\mathrm{O}_{2}$. (o) Calculate the maximum mass of NO that could be obtained by burning $1.00 \mathrm{kg}$ of $\mathrm{NH}_{3}$ and the mass of oxygen required. (f) In practice, $1.45 \mathrm{kg}$ of $\mathrm{NO}$ were obtained from the reaction in (e). What was the percentage yield of the reaction?

The fertilizer, ammonium nitrate, is made by reacting ammonia with nitric acid. (Section $1.4)$
(a) Write a balanced equation, with state symbols, for the reaction of ammonia gas with nitric acid to form a solution of ammonium nitrate.
(b) Rewrite the equation to show the ions present in the reactants and products, and hence write an ionic equation for the reaction.
(c) What type of chemical reaction does your equation represent?
Nitric acid is manufactured from ammonia. The first stage in this process involves burning ammonia in oxygen on the surface of a platinum gauze catalyst. The products are $\mathrm{NO}$ and $\mathrm{H}_{2} \mathrm{O}$. (d) Construct a balanced equation for the buming of $\mathrm{NH}_{3}$ in $\mathrm{O}_{2}$.
(o) Calculate the maximum mass of NO that could be obtained by burning $1.00 \mathrm{kg}$ of $\mathrm{NH}_{3}$ and the mass of oxygen required.
(f) In practice, $1.45 \mathrm{kg}$ of $\mathrm{NO}$ were obtained from the reaction in
(e). What was the percentage yield of the reaction?

Key Concepts

Balanced Chemical Equations

A balanced chemical equation ensures that the number of atoms for each element is the same on both sides of the reaction, adhering to the law of conservation of mass. It also includes state symbols to describe the physical state of each substance involved. This concept is fundamental in chemistry for accurately representing chemical reactions and for performing stoichiometric calculations.

Ionic Equations

Ionic equations break down aqueous compounds into their constituent ions, showing only the species that actively participate in the chemical reaction. This form of representation helps to identify spectator ions and provides a clearer picture of the actual chemical change, especially in reactions occurring in solution.

Acid-Base Neutralization Reactions

Acid-base reactions involve the transfer of a proton (H?) from an acid to a base, resulting in the formation of water and a salt. The reaction between ammonia, a base, and nitric acid, an acid, forming ammonium nitrate is a classic example of this type of reaction. Recognizing and classifying these reactions is key in understanding solution chemistry and synthesis.

Oxidation Reactions (Combustion of Ammonia)

Oxidation reactions are those in which a substance loses electrons, often involving oxygen. The burning of ammonia in oxygen, typically in the presence of a platinum catalyst to produce nitric oxide and water, is an example of an oxidation (or combustion) reaction. Balancing such equations requires careful consideration of the redox changes and stoichiometric coefficients.

Stoichiometry

Stoichiometry involves quantitative relationships between reactants and products in a chemical reaction. It is used to calculate the amount of reactants needed, predict the quantities of products formed, and determine limiting reactants. This concept is critical when determining yields and scaling reactions from the laboratory to industrial production.

Percentage Yield

Percentage yield measures the efficiency of a reaction by comparing the actual yield obtained from an experiment to the theoretical yield predicted by stoichiometry. This concept helps chemists assess the practicality and efficiency of chemical processes and to identify potential areas for optimization in reaction conditions.

Transcript

0:00 All right.

00:01 For this problem, we're going to be covering balancing equations, writing ionic equations, types of reactions, finding maximum mass, and percent yield.

00:12 This is the equation that we're given.

00:14 So the first thing we're going to want to do is balance this equation.

00:18 And the way i like to do that is first i'm going to look for polyatomic ions that appear on both sides of the equation.

00:24 And i see no3 on both sides of the equation.

00:28 What that means is that means we can treat those as one component.

00:31 And not break it up and have to do all kinds of other math for that.

00:36 So we're going to put the elements and components that we have over here first.

00:42 So n, we have h, and then we have our no3 component.

00:49 All right.

00:49 Next thing i'm going to do is i'm going to add all the components of these elements that we have.

00:54 So we have one nitrogen on this side, four hydrogens on this side, and one n -o -3.

01:03 Same thing on the other side.

01:04 We're going to add it up, 1, 4, and 1.

01:08 So, that's already balanced.

01:13 The next thing we're going to do is we're going to find the ionic equation, which basically means we're going to take all of these components, we're going to break them up into their individual charges, find out which ones appear on both sides so we can cancel it, and then look at what the reaction is really doing.

01:29 So we can't break up this polyatomic ion here.

01:33 So we're going to put nh3, which has a neutral charge, we can break up this into its hydrogen and no3 components, so we're going to have plus hydrogen, which has a positive 1 charge, plus n03, which has a negative 1 charge.

02:03 All right, and then that's going to yield.

02:06 We can break this up to nh4, which has a positive 1 charge, and we can add n03 with its negative 1 charge.

02:21 All right, so that's what we call our complete ionic equation.

02:31 We're going to also look at the net ionic equation, which is what happens after we remove our spectators or things that appear on both sides of the equation and therefore cancel each other out.

02:42 So the only thing i see here that'll cancel out is n -o -3, since that appears on both sides.

02:50 So our net ionic reaction is going to be, sorry, equation, not reaction, is going to be nh3, plus h yields nh4.

03:08 So in essence, this is what this reaction boils down to.

03:15 All right, the next thing we're going to do is we're going to figure out what type of reaction this is.

03:20 Now, a big hint is that this molecule here is called nitric acid.

03:32 So really, that's an acid.

03:35 So that says to me this is probably an acid -based reaction.

03:38 But another way you can verify that is you can look at this.

03:41 Hydrogen here and we can notice that this hydrogen is going to get transferred because it winds up here so it's no longer going to be a part of this n -o -3 it's now a part of this nh4 so that's a hydrogen transfer which indicates acid -based reaction.

04:05 All right the next thing we're going to do is we're given this other equation that we're going to look at which is nh3 plus o2 yields no and o plus h2o.

04:26 And i've spaced this out because i already know that this balancing thing is going to be a lot more convoluted than the one that we had up here.

04:33 There are no polyatomic ions that we can keep together, so we just have to list each individual element.

04:38 So we've got nitrogen, hydrogen, and oxygen.

04:42 I'm going to go through these, and i'm going to add up what we've got.

04:45 One nitrogen, three hydrogens, two oxygens.

04:51 On the other side, one nitrogen, two hydrogen, two hydrogen, two hydrogen, and two oxygens.

05:00 Unfortunately, this is not balanced, and it's going to take a little while and a lot of just working things out and seeing what works.

05:11 So the first thing i'm going to do is i'm going to try to balance these hydrogens by saying we have two of these here and three of these here.

05:20 Well, now i have to update everything.

05:22 So on this side, we have now two nitrogen and six hydrogens.

05:28 On this side, we now have six hydrogens and four oxygens.

05:38 As a general rule of thumb, when you're doing things like this, don't mess with the oxygen.

05:42 It'll probably balance itself out, and if not, it'll have a really easy thing right at the end.

05:47 But don't mess with that right now.

05:50 Well, this is all well and good, except we don't have our nitrogen's balanced or our oxygen's balance anymore.

05:56 So i'm going to update this over here and give this one to nitrogen.

06:02 Well, that throws off our oxygens too, so now we have five of these, which is really inconvenient because if it was an even number, we could just put whatever we need to multiply by two here, and it would be balanced, but it's not, it's an odd number.

06:16 So this is what i mean when i say trying things and seeing if they work...

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