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Which is the limiting reactant when 50 mn with 2.75 les of copper are reacted moles 0f silver nitrate In the follawing Cu equation? ZAgNOa Cu(NO,hz - 2A8AgNOjCu(NOz...

Question

Which is the limiting reactant when 50 mn with 2.75 les of copper are reacted moles 0f silver nitrate In the follawing Cu equation? ZAgNOa Cu(NO,hz - 2A8AgNOjCu(NOzhNext

Which is the limiting reactant when 50 mn with 2.75 les of copper are reacted moles 0f silver nitrate In the follawing Cu equation? ZAgNOa Cu(NO,hz - 2A8 AgNOj Cu(NOzh Next



Answers

a. If 2.50 mol of copper and 5.50 mol of silver nitrate are available to react by single replacement, identify the limiting reactant.
b. Determine the amount in moles of excess reactant remaining.
c. Determine the amount in moles of each product formed.
d. Determine the mass of each product formed.

According to the reaction given how many grams of copper to nitrate will be formed upon the complete reaction of 26.8 g of copper with excess silver nitrate starting from 26.8 g of copper. Convert to moles using the Moller Mouse Reactions to Okayama Tree one mole of copper, 21 mole of copper nitrate and come right back to grams. One mole of copper nitrate has a mole amounts of 187.86 Um, 0.6 g. This works out to 79.1 g of copper nitrate.

There are multiple steps involved in each part of this question. First, if we have five g of each reactant, we need to figure out which reactant is the limiting reactant, Then knowing which one is the limiting reactant, we can use its amount to calculate the amount of products produced. But before we can do any of that, we need to have a balanced chemical reaction. The first chemical reaction is sodium reacting with bromine, BR two producing sodium bromide. And as is it's not balanced. We have to grow means we only have one bro. Mean? So we need to put it to here. Now we have two bro means and we just introduced to sodium. So we need to put it to here. Now it's balanced. So let's start with five g of sodium and see how much sodium bromide can be created. If all 5g react five g of sodium could be converted into moles of sodium by divided by the molar mass of sodium. Then when we have mold sodium we can convert it to moles of sodium bromide, recognizing it's a 2-2 more relationship. Then we have moles sodium bromide. We can convert to gram sodium bromide by simply multiplying by the molar mass of sodium bromide And we get 22.4 g sodium bromide. If all of the sodium reacted, the second reactant is brimming. So if we have five g roaming we can convert it to moles bro. Mean by divided by the molar mass bro. Main, Then we'll go from moles roaming to mold sodium bromide, recognizing that it's a 1-2 more relationship from the balanced chemical reaction. Then we have moles sodium bromide will multiply by the molar mass sodium bromide as we did appear to calculate the grams of sodium bromide that would be produced if all five g of bromine reacted And we get 6.44 g sodium bromide. So the limiting reactant is always the one that produces the least amount of product. So bro Main five g roaming produced the least amount of product. Therefore it is the limiting reactant And the actual amount of sodium bromide that would be produced would be the amount created from it. 6.44 g. Now for part B, the reaction is zinc reacting with copper, two sulphate producing copper and zinc to sulphate. And as is we have one sink, one zinc, one copper, one copper, one sulfate, one sulfate. So it's already balanced. Then we'll start with five g inc see how much product we can make. You can choose rich product. We need to calculate the amount for both of these. So just choose one of them to identify the limiting reactant. The limiting reactant will always produce the least amount of both products. So I'm going to convert zinc into molds inc by dividing by its smaller mass, Then go from old sink into. I'm going to choose copper molds copper, which is a 1 to 1. Strike geometric relationship. Then when I have moles copper all multiplied by the molar mass. Copper take it to graham's copper. If all think all five g reacts completely, We'll get 4.89 g copper. But what about the five g copper two sulphate. Let's convert the grams copper two sulphate into molds Copper two sulphate by divided by the molar mass. Copper two sulphate. Then we'll convert the molds. Copper two sulphate into molds copper A 1- one relationship. Then, when we have moles, copper will multiply by the molar mass copper to get the grams copper and we get 1.99 g copper being produced from the five g copper two sulphate. So copper two sulphate producing the least amount of copper is the limiting reactant. And the actual amount of copper produced would be the 1.99 g coming from the copper two sulphate. Now, knowing copper two sulphate is the limiting reactant. We can use its amount to calculate the amount of the second product, zinc sulfate, Take five g copper two sulphate converted into molds copper two sulphate by divided by the molar mass, Then convert the molds, copper two sulphate into moles of zinc sulfate. Knowing it's a 1 to 1 relationship, one to one and then look up or calculate the molar mass of zinc sulfate, multiplied by the molar mass zinc sulfate to get the grams zinc sulfate And we get 5.06 g zinc sulfate being created with the 1.99 g copper. Now, for part C, the chemical reaction is ammonium chloride reacting with sodium hydroxide producing ammonia, gas, water and sodium chloride. This one actually has three products for which we need to calculate the masses. But before we can do that, let's balance the chemical reaction and identify the limiting reactant to balance the chemical reaction to recognize that we have one nitrogen, 1 nitrogen five hydrogen, five hydrogen, one chlorine, one chlorine, one sodium, 1 sodium and one oxygen and one oxygen. So we're balanced. So let's start with the five g ammonium chloride. If all of it is consumed, we can convert its grams into molds by divided by the molar mass ammonium chloride. Then recognize that it is a one To one relationship to the product Ammonia. I'm going to choose ammonia. In order to identify the limiting reactant, you could choose any product you want. The limiting reactant will produce the least amount of any product that is chosen When I have moles of Ammonia, I can convert two g of ammonia by multiplying by its smaller mass. If all five g ammonium chloride is consumed will get 1.59 g of ammonia. But we also have five g of sodium hydroxide. So let's convert the grand sodium hydroxide into ml sodium hydroxide by dividing by its smaller mass. Then we'll go from moles sodium hydroxide, two moles of the same product that was chosen ammonia. Then when we have moles, ammonia will multiply by the molar mass ammonia to get Graham's ammonia 2.13 g pneumonia. So because the ammonium chloride produces the least amount of ammonia, Then it is the limiting reactant and the mass of Ammonia that is created comes from it 1.59 g. Now we know now knowing the ammonium chloride is the limiting reactant, we can use it to calculate the mass of the other two additional products. We'll start again with the five g ammonium chloride converted into molds, ammonium chloride by dividing by its smaller mass. Then we'll go from moles ammonium chloride, two moles of water, which is 1-1. Then knowing moles of water, we can multiply by the molar mass of water to get grams water and we get 1.68 g water. Then for the third product will do the same thing. Start the same way With a five g ammonium chloride divided by the molar mass ammonium chloride to get moles ammonium chloride, Recognise the 1-1 relationship between ammonium chloride and sodium chloride and then multiply by the molar mass sodium chloride to get grams sodium chloride, We get 5.46 g sodium chloride. These being the masses of product created recognizing ammonium chloride is the limiting reacted Now for the last one part D It's iron three oxide reacting with carbon monoxide producing iron metal and carbon dioxide. To balance this will put a three here. I know we don't have three carbons and we don't have six oxygen's but we do have We do need to come up with a multiple of three and 2 Or a number that is divisible by three and 2 and that would be six. So we put a three there to get six oxygen's. They also gave us three carbons. Will put a three there to get three carbons. We have two irons here, so we'll put it to their And now let's verify that the oxygen's are balanced 123123 that's six and then three times two is six. So it's balanced. So let's start with the five g iron three oxide converted to moles iron three oxide. By dividing by the molar mass iron three oxide. Then we'll go from moles iron three oxide, two moles of either product. To identify the limiting reactant. You choose one, I'll choose iron. One more iron three oxide produces two moles of iron. Now that we have moles iron, we can multiply by the molar mass of iron to get grams of iron. If all five g of iron three oxide were consumed would produce 3.50 g of iron. But we also have five g of carbon monoxide. So let's divide that by its smaller mass to get molds carbon monoxide, Then we'll go from moles carbon monoxide to most of the same product iron with a 2-3 relationship. Then we have moles of iron. Well multiply by the molar mass as we did up above to get grams of iron. We see that the smaller amount is 3.5 g iron coming from five g iron three oxide. So the Iron three oxide is the limiting reactant and the actual amount of iron produced is the 3.50 g. Now, knowing iron three oxide is the limiting reactant, we can use its mass to calculate the mass of the second product carbon dioxide, we'll go from five g iron three oxide, two moles by dividing by its smaller mass. Then we'll see the 1, 2. Three relationships Between Iron three oxide and carbon dioxide. Then we have moles. Carbon dioxide will multiply by its molar mass to get grams carbon dioxide and we get 4.13g carbon dioxide.

So in this problem, we want to know how much copper nitrate is produced if we also produce 2.25 grams of silver. So the first thing I did was kind of right out a chemical reaction for the word equation that were given in the problem. And I also went to the periodic table and got some molar masses that we're gonna need. So now let's start with 2.25 grams of silver. So we have here, um, the molar mass of silver, which we're gonna use to convert this two moles. So 107 0.87 grams of silver is ah, mole of silver. And we also have the mole ratio Two moles of silver from our Sochi. A metric coefficients is equal. Teoh one mole of copper night trade. And we also have the molar mass of copper nitrate. So a little of copper nitrate is equal to 187.56 grams. So now it's multiply all that out. We have 2.25 times. Ah 100 in 87.56 divided by 107 0.87 Divided by two and we have 1.96 grams of copper nitrate. What we want to do next is a really similar process for each of the reactant. So let's start with copper. We still convert two moles of silver and then we have the mole ratio of one mole of copper equals two moles of silver, and we use the more massive copper to convert we have one hole of copper is 63.55 grams and we're gonna multiply this out to get 2.25 times 63.55 divided by 107.87 divided by two and we have 0.66 grams of copper. Next, we repeat the process for our silver nitrate. So we have two moles. Uh, silver is equal to two moles of silver nitrate, and we know that from our Miller Mass. One bowl of silver nitrate is equal to 169.787 grams. So now we just have to multiply all this together. 2.25 times two times 169.87 divided by 107.87 divided by two. That gives us 3.54 grams of silver nitrate

We have a reaction between Cooper has ever nitrates. So between these two were at in every way every shoulder produce silver and unknown reacting. So we are going to complete this chemical reaction. And also far the mass off the unknown reacting and also the unknown. Reactive. Okay, so you can see that this is essentially a single displacement reaction. So that means helping in pace the re agent first. So it will cooperate. We we pay silver. Um, we have super over here. And also you will become copper. No treat copper nitrate. Okay, The next step, we're going to bars the chemical reaction because we have to nature under right. So we have to put two nitrate. I mean two infernal super nitrates. And also we have two infernal silver. Okay, so we have to a chemical reaction. And also you filed that the unknown part of will be covered nitrate. So that's that we're going to found a mass off couple in a nitrate. Okay, so we know that the mass off copper he is 65 on five. As if in nitrates, leafy 9.8 scram. So the total mass will be adding both to get her and you can just soaked elevator and you should be able to find out. It would be slow for you wonderfully, Graham. So you just need to add this to together. We have found the total mass of Cooper and super nitrates. Okay, so by conservation of mass, the mass, the total mass of the reaction should not be changing because we don't have any reactions. Are leaving the waiting. We're back. So for several gas can be can be the best. So so But for this game, we don't have a solid and those over increase solution. So the master to the master be the same. So we're given the massive silver is 215.8 gram. So the mass off couple naturally must be equals to our total mess. Sub trick our mass off silver and you should be able to find that the massive silver everybody goes to 187 put a fight crime and disa coded different by a total mess. It's a trick. Our super mass and then we will have the mass of supernatural is 187.5


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