Skip to main content
Pearson+ LogoPearson+ Logo
Ch.7 - Periodic Properties of the Elements
Back
Previous problem
Next problem
All textbooksBrown 14th EditionCh.7 - Periodic Properties of the ElementsProblem 113c

Chapter 7, Problem 113c

When magnesium metal is burned in air (Figure 3.6), two products are produced. One is magnesium oxide, MgO. The other is the product of the reaction of Mg with molecular nitrogen, magnesium nitride. When water is added to magnesium nitride, it reacts to form magnesium oxide and ammonia gas. (c) In an experiment, a piece of magnesium ribbon is burned in air in a crucible. The mass of the mixture of MgO and magnesium nitride after burning is 0.470 g. Water is added to the crucible, further reaction occurs, and the crucible is heated to dryness until the final product is 0.486 g of MgO. What was the mass percentage of magnesium nitride in the mixture obtained after the initial burning?

Verified Solution
Video duration:
17m
This video solution was recommended by our tutors as helpful for the problem above.
1978
views
Was this helpful?

Video transcript

Hi everyone for this problem. It reads sodium can be reacted with oxygen at 130-200°C to produce sodium oxide and sodium peroxide, sodium oxide can further absorb oxygen to produce sodium peroxide. A piece of sodium metal reacted with oxygen and produced a mixture of 0.496 g of white and yellow solid. The mixture was further exposed to oxygen until the final product is pure yellow solid with a mass of 0.5 to five g, calculate the mass percentage of sodium oxide in the initial mixture. So, our goal here is to calculate the mass percentage of our sodium oxide. Alright, and let's recall that mass percentage is mass of solute over mass of the solution Times 100%. So in this case, our salute is going to be the sodium oxide. Saw grams of sodium oxide and this is going to be over our mass of the solution, which we're told in the problem. The mass of the mixture is 0.496 g. So we know what the mass of the solution is. So now we need to figure out what is our mass of sodium oxide. So we can calculate the mass percentage. Okay, so let's go ahead and start off by writing out our mixtures. Okay, So we have the following and the following. Okay, So after the first mixture was exposed to oxygen, all of the sodium is converted to sodium peroxide. So what we're going to do here is we're going to start off by calculating the initial mass of sodium and we're going to calculate the initial mass of sodium using the given the initial given for our sodium peroxide. And that given was 0.5 to five g of sodium peroxide. So we want to go from grams of sodium peroxide, two g of sodium. Okay, so let's go ahead and get started. So We want our sodium peroxide to cancel. So that means we're going to need the molar mass of sodium peroxide. So one mole of sodium peroxide using our periodic table, it is 77.98 g of sodium peroxide. So looking at our units, we see our grams of sodium peroxide cancel. Remember we're trying to go from grams of sodium peroxide, two g of sodium. So now what we can do is use the multiple ratio of sodium peroxide and sodium. So we see here that are multiple ratio is for every two moles of sodium consumed, there is one mole of sodium peroxide produced. So that's our multiple ratio. Let's go ahead and plug that in. So for every two moles of sodium consumed, There is one mole of sodium peroxide produce. So we want this sodium peroxide to cancel. And it does. So now we have moles of sodium. And remember we want grams of sodium. So we're going to use the molar mass of sodium Now in one mole of sodium, There is 22.99 g of sodium. Alright, so now our moles of sodium cancel. And we're left with grams of sodium, which is what we want for this initial step. So, once we do that calculation, we get 0.3096g of sodium. Alright, so now that we know are massive sodium, what we're going to do here is define some variables. Because our end goal is to find our mass of sodium oxide. Alright, so our variables that we're going to define is x. Is going to equal our grams of sodium converted to sodium peroxide. Why is going to equal grams of sodium converted to sodium oxide? And we're going to say X. Is also going to equal this Initial mass of sodium. We just calculated so .3096 g -Y. So these are three variables. And let's just go ahead and take a note of all of our molar masses. So our molar mass of sodium peroxide is 77.98 g per mole molar mass of sodium oxide is 61.98 grams per mole. And moller mass of sodium is 22.99 g per mole. So we're just going to write this year so we can reference it as we continue this problem. So the first thing we're going to do is we're going to figure out our grams of sodium peroxide using the variables we just defined. So our grams of sodium peroxide. Okay, it's going to equal X grams of sodium. Okay, so that's this one. Alright, so we're going to start there. So we have our grams of sodium peroxide is equal to x grams of sodium. So now we're going to do the the dimensional analysis. So we have in one mole of sodium There is 22.99 g of sodium. Okay, So our grams of sodium cancel and we're left in moles of sodium. Now I'm using the multiple ratio and one mole of sodium Peroxide. There is two moles of sodium. So here our moles of sodium cancel. And in one mole and one mole of sodium peroxide, There is 77.98 g of sodium peroxide. Alright, so this cancels. So now we're left with grams of sodium peroxide. So this X grams of sodium, we're going to replace it with how we define the variable. So we define X as this one right here. Okay, so we're going to go ahead and substitute it. Okay, So we have zero point Or let me write it in a different colour. So we're substituting how we define this variable, and we define it as 0.3096 grams minus y. Now we're going to do the math calculation for the rest of it. So the rest of it, once we do the calculation, so we're copying this is the same. Okay, so we're going to do this math right here and once we do that, the value we get is 1.6959. Alright, so our grams of sodium peroxide equals this. Alright, so this is step one. So we know this. Now, we're going to do the same thing. We're going to do the same thing, but for grams of sodium oxide. So let's do that in a different color. So our grams of sodium oxide is equal to the way we defined it is why grams of sodium. Alright, So, during our dimensional analysis and one mole of sodium, there is 22.99 g of sodium. in two moles of sodium oxide, There is four moles of sodium and in one mole of sodium oxide There is 61.98 grams of sodium oxide. Okay, so looking at our units, we see the cancel. And we're left with grams of sodium oxide. Okay, So we're going to do the same thing we just did above. So our grams of sodium oxide is going to equal why times What we get for this calculation here. Okay, and that calculation, once we do it, we get 1.3480. Alright, so we're paying attention to these two here. And how we define them. So our grams of sodium oxide equals the following. And our grams of our grams of sodium peroxide equals the following. And our grams of sodium oxide equals the following. Alright, So now, what we're going to do is solve for y Okay, and the way we're going to do that is by writing out a equation And our equation is going to be so we know we have in the problem we're told we have .496g of the mixture. Okay, so we're going to say 0.496 which is our mixture is equal to the two that we just solved for, Add it up. Okay, so we have 0.3096 -6. Times 1.659 plus 1.3480 Y. So the two variables we just defined, we set them equal to the mass of our mixture. The sum of these two equals the mass of our mixture. That's what we just did. And now we're going to solve for Y. So let's go ahead and simplify this. Alright, so we have 0.496 is equal to and we're gonna foil this out. Okay, so we have 0.5250 -1.6 959 Y plus 1.3480. Y. Alright, so simplifying this more, we get 0.3479 Y. Is equal to 0.29. Alright. And we're solving for y. So we're going to divide both sides by 0.3479. And when we do that we get y is equal to 0.8336 g of N. A. Or grams of sodium and this is our grams of sodium that is converted to sodium oxide. So now, because remember we said in the beginning, after the first mixture was exposed to oxygen, all of the sodium is converted to sodium peroxide. Okay, so this is our grams of sodium oxide. So now that we know our grams of sodium, we can go ahead and calculate our grams of sodium oxide. Okay, we're almost finished. So we have we're going to use the value. We just calculated. So 0.8336 g of sodium. And we want to go from grams of sodium two g of sodium Oxide. So we're going to use our dimensional analysis here. So we want to get rid of grams of sodium and go to moles. So in one mole of sodium there is 22.99 g of sodium. So now our grams cancel. Now we want to go from moles of sodium two moles of sodium oxide. We'll use our multiple ratio. So there's two moles of sodium Oxide for every four moles of sodium. Okay, so moles cancel. Now we want to go from moles of sodium oxide, two g of sodium oxide and we'll use molar mass in one mole of sodium oxide. There is 61.98 g of sodium oxide. So our moles of sodium oxide cancel and we have grams of sodium oxide, which gives us a final answer of 0.11-4 g of sodium oxide. So remember if we go back to the top. Our goal was to find our grams of sodium oxide to do the mass percent. So now we can go ahead and plug in. So we knew the grams of the initial mixture was 0.496 g. And now we know the grams of the mass of the sol U. Which we just saw four is 0.11 to four g. And this is going to be times 100%. So Our final answer is going to be 22.7%. This is the mass percentage of sodium oxide in the initial mixture. That's it for this problem. I hope this was helpful.
Previous problemNext problem
Related Practice
Textbook Question

One way to measure ionization energies is ultraviolet photoelectron spectroscopy (PES), a technique based on the photoelectric effect. (Section 6.2) In PES, monochromatic light is directed onto a sample, causing electrons to be emitted. The kinetic energy of the emitted electrons is measured. The difference between the energy of the photons and the kinetic energy of the electrons corresponds to the energy needed to remove the electrons (that is, the ionization energy). Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength 58.4 nm. (c) The kinetic energy of the emitted electrons is measured to be 1.72 * 10-18 J. What is the first ionization energy of Hg, in kJ>mol?

560
views
Textbook Question

One way to measure ionization energies is ultraviolet photoelectron spectroscopy (PES), a technique based on the photoelectric effect. (Section 6.2) In PES, monochromatic light is directed onto a sample, causing electrons to be emitted. The kinetic energy of the emitted electrons is measured. The difference between the energy of the photons and the kinetic energy of the electrons corresponds to the energy needed to remove the electrons (that is, the ionization energy). Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength 58.4 nm. (d) Using Figure 7.10, determine which of the halogen elements has a first ionization energy closest to that of mercury.

990
views
Textbook Question
Mercury in the environment can exist in oxidation states 0, +1, and +2. One major question in environmental chemistry research is how to best measure the oxidation state of mercury in natural systems; this is made more complicated by the fact that mercury can be reduced or oxidized on surfaces differently than it would be if it were free in solution. XPS, X-ray photoelectron spectroscopy, is a technique related to PES (see Exercise 7.111), but instead of using ultraviolet light to eject valence electrons, X rays are used to eject core electrons. The energies of the core electrons are different for different oxidation states of the element. In one set of experiments, researchers examined mercury contamination of minerals in water. They measured the XPS signals that corresponded to electrons ejected from mercury's 4f orbitals at 105 eV, from an X-ray source that provided 1253.6 eV of energy 11 ev = 1.602 * 10-19J2. The oxygen on the mineral surface gave emitted electron energies at 531 eV, corresponding to the 1s orbital of oxygen. Overall the researchers concluded that oxidation states were +2 for Hg and -2 for O. (b) Compare the energies of the 4f electrons in mercury and the 1s electrons in oxygen from these data to the first ionization energies of mercury and oxygen from the data in this chapter.
2254
views
Textbook Question

When magnesium metal is burned in air (Figure 3.6), two products are produced. One is magnesium oxide, MgO. The other is the product of the reaction of Mg with molecular nitrogen, magnesium nitride. When water is added to magnesium nitride, it reacts to form magnesium oxide and ammonia gas. (d) Magnesium nitride can also be formed by reaction of the metal with ammonia at high temperature. Write a balanced equation for this reaction. If a 6.3-g Mg ribbon reacts with 2.57 g NH31g2 and the reaction goes to completion, which component is the limiting reactant? What mass of H21g2 is formed in the reaction?

1898
views
Textbook Question

Potassium superoxide, KO2, is often used in oxygen masks (such as those used by firefighters) because KO2 reacts with CO2 to release molecular oxygen. Experiments indicate that 2 mol of KO2(s) react with each mole of CO2(g). (c) What mass of KO2(s) is needed to consume 18.0 g CO2(g)? What mass of O2(g) is produced during this reaction?

1593
views