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Determine the number of moles and the mass requested for each reaction in Exercise 4.44

(a) $0.0686 \mathrm{mol} \mathrm{Mg}, 1.67 \mathrm{g} \mathrm{Mg}$(b) $2.701 \times 10^{-3} \mathrm{mol} \mathrm{O}_{2}, 0.08643 \mathrm{g} \mathrm{O}_{2}$(c) $6.43 \mathrm{mol} \mathrm{MgCO}_{3}, 542 \mathrm{g} \mathrm{MgCO}_{3}$(d) $768 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}, 13.8 \mathrm{kg} \mathrm{H}_{2} \mathrm{O}$(e) $16.31 \mathrm{mol} \mathrm{BaO}_{2}, 2762 \mathrm{g} \mathrm{BaO}_{2}$(f) $0.207 \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{4}, 65.81 \mathrm{g} \mathrm{C}_{2} \mathrm{H}_{4}$

Chemistry 101

Chapter 4

Stoichiometry of Chemical Reactions

Chemical reactions and Stoichiometry

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Hello everyone today. We're going to solve for requested moles and mass and specific problems and exercise. 4.44. Um Let's go ahead and get started. We're going to use a technique called conversion factors. So let's start with what we already know. We know we have five g of hcl. So let's start with five g Hcl for conversion factors. You need to put whatever is on top here, on the bottom down here almost like a fraction. So we're gonna use molar mass to be able to do this in the molar mass of hcl is 36 point four or 5 a day. Okay. But grams per one mole now we need to do this because we need to be able to get a multiple ratio and we can see here that for every two moles of H C. O. We have one mole of MG. Which is actually what we're looking um for the molds and mask for. So for every two moles hcl we have one more N. G. Now to get the smalls of MG. We need to divide five by 36.458 And then divide that number by two. You can multiply by one if you want. But since it's going to be the same number, you don't have to. And this is going to give us an answer of 10.6 85 72 moles of MG. And now we have the first answer to our problem. We're gonna go ahead and use the answer. We just got to solve for the second part. So 720.68572 malls MG. Okay put that in a conversion factor and we will do a More mass conversion. Once again the molar mass of Magnesium is 24.305 grams. Mhm. Mhm. And then you multiply these two numbers on top here. So we take .06857. To multiply that by 24.305g. And we get our answer of 166 g. My museum. And now we have both of our answers. So let's go ahead and move on to our next question which is finding diatonic oxygen's moles and mass in this specific equation. So let's start with what we know. We know we have 1.252 g of silver, one oxide molar mass conversion. We need to be able to do a multiple ratio. So the molar mass of This compound is to 31.735 grounds. Mhm Per one more. Let's go ahead and do multiple ratio. We can see again that for every uh more of a G two, we have two for every one of our die atomic oxygen. So we need to put two more AG two L for one more die atomic oxygen. And then we're going to divide 1252 divided by to 31.735 g to get our moles of silver one oxide. And then convert those moles by dividing by two to get one mole of die atomic oxygen. And we get an answer uh 2.701 times 10 to the negative third power. Mhm. Mhm. And then once again we're going to use this number. Put it in another conversion factor which is going to give us our final answer of grams of die atomic oxygen. So do you use molar mass? But Okay. And the molar mass of 02 is going to be 32 g mm. So we multiply these two numbers because they're on top. So we take 2.701 times 10 to the negative third power times 32 grounds. And we get our answer of point 08644 g 02. And now we have both of our answers and we have solved our second problem. Yeah. All right. Moving on to our 3rd problem. We're looking for a magnesium carbonate. We already know that carbon dioxide has in this reaction um has produced 238 283 g of that. So let's start with what we know 283 g of C. 02 What's do more mass conversion. So we can do mobile mole ratio. So 283 g. Let's do 44.01 grams and one more C 02 Then we're going to move on to one more C. 02 for every one more of mg C 03 And we're going to divide these two numbers And that will give us 6.43 moles of magnesium carbonate. Once again we're going to take this number and put it back into another converting factor so we can find our mass. Let's do one more. Excuse me, one More. And do you see your three and then our Mueller mass for That is 84.3139 84 3139 grounds. Mhm. Yeah. And that gives us an answer. 542 g magnesium carbonate. And we have a final part of our answer. No, we can move on to our fourth problem and we're looking for the malls and mass of water in this combustion reaction and we know that we have 20 kg of C two H two. However, all of our unit measurements are ingram's as we can see here. So to be able to use conversion factors, we're going to have to convert this first handedly two g so that it works. And the equation so 20 Kg, we know that for every one kg There is 1000 g and this gives us 20 1000 g equivalent. All right, once again, let's start with what we know and we know that we have 20,000 g of C2 H two to start off with. We're going to do Our conversion Doctor of Montana's. So 26.04 grams per one more. So you two H 2. Then we do to morals of C two H two for every two moles of H 20 So again, another 1 to 1 mole ratio. You don't have to include this step, but it is encouraged to go ahead and just get some more practice with multiplying and dividing Someone to Normalcy to H. two. For every jewel h. Okay. 20,000 divided by 26.04 times two divided by two. We get a super awesome answer of 768 point 049 right now. Just like we did before. We're going to take that answer. Let's go ahead and put a label on that labels are shape are important. 7 68.049. Mhm H 20 For every one mole Mhm Of H. 20. You will get 18.8015- eight g of that compound. And that is going to Give you a really big answer. And then you can go ahead and divide that again by 1000. So that you can um Write it in kg if you want to. You don't have to. Since everything else is already in g. We get 13 836.6779 So if we divide that by 1000 we essentially get yeah, 13.8 kilograms of age too. And then we have our mass and we also have our moles. So that's the answer to part four. Let's move on to our fifth problem. And here we know that we have 2.500 kg. Again, we're gonna convert that out. So 2.500 kg. Mhm. For every one kg we have 1000 grams. Which is going to give us 2500 ground of boko. Mhm. Now we just use that number that we have. Yeah. And start our conversion factors. So 2500 g B. O. The mhm. And the molar mass of the i. is 1 53.33. Okay. Everyone more. Mhm. And then yeah, we take our multiple ratio. We can see that we have Actually let's stop there. Let's make our more another conversion factor. Since it's a multiple ratio, I'm going to give you an example of what I mean by stopping. So we can see that we have to to to and that means that for every two moles of BA 02 We have two moles of video which is sent which is essentially a coal equivalent to a 1-1 ratio as well. So we don't have to multiply any further. And we're going to take 2500 Divided by 1 33.33g. And we're going to get an answer of This needs to be a two. We're Then multiplied by two morals of B. O. So 2500 divided by 153.33 times two. And we get 16.30 old B. O. Or you can write it out if you want. I'll go ahead and show you what that looks like. Um You're just right to more and then two. Mhm. Mhm. And You get the same answer. Sometimes it's just easier to write everything else so that you can actually visualize it, visualize it. But your answer is going to be 16.30 more B area oh two. Now we're going to use this number and a conversion factor. So 16.30 mm. Be a Po two. And the more massive vo two is 1 69.33. Mhm. Mhm. Mhm. For one more of it. Excuse me. I read that on the bottom. Remember that you're the same unit of measure on top. Goes on bottom. So we need one mold down here and we need 160 9 23 3 g on top to give us Mhm 27 60079 g BA 02. And we have solved those parts of the problem and we have finished our fifth problems. So let's go ahead and move on to our final problem. Once again, we're going to start with what we know. We can see that we have 955g. So 9.55 g. don't need to do any extra conversions there. With that unit of measurement. We can jump straight into our Mueller mass conversion. The molar mass of this compound is 46.07 g. Then one more Uh see to age 5 to age. Yeah. And since this is a mole to mole ratio one more of this of our product we're reacting to in this case that we are finding um for this one there is no Coefficient in front of it. So we can just go ahead and stop their divide out. Which we get .2073. So we need to take 9.55, divided by 46.07 to get .2073 moles AC two H 4. Okay. And then we just take that number back into another conversion factor. Mhm. Mhm. For more mass. And we're going to get Mhm. Yeah. The grounds 28.05. So you need to take .2073 multiply that by 28.05. And you're going to get 5.81 g of c two age four. And you have your final answer. And that is it for this problem set. Thank you so much for watching.

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