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Question 30pes;If you have 1Og of Nzand 1Og of Ozhow many grams of excess reactant will reaction is complete? remain when theN2 + Cz > ZNO2.750 1250* 100114...

Question

Question 30pes;If you have 1Og of Nzand 1Og of Ozhow many grams of excess reactant will reaction is complete? remain when theN2 + Cz > ZNO2.750 1250* 100114

Question 30 pes; If you have 1Og of Nzand 1Og of Ozhow many grams of excess reactant will reaction is complete? remain when the N2 + Cz > ZNO 2.75 0 125 0* 10 0114



Answers

If a mixture of $100 . \mathrm{g} \mathrm{Al}$ and $200 . \mathrm{g} \mathrm{MnO}$ is reacted $\mathrm{ac}$ cording to the reaction $$ 2 \mathrm{Al}(\mathrm{s})+3 \mathrm{MnO}(\mathrm{s}) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{Mn}(\mathrm{s}) $$ which reactant is in excess and how many grams of it remain when the reaction is complete?

Starting from 4.2 moles, uh, zinc sulfide and six points eat smalls oxygen. Let's so for the amount of excess reactant that remains. We're going Teoh, use our story geometry. I'm gonna pick a product. The product I'll pick zinc oxide. So two moles of think so far to two moles of zinc oxide would yield 4.2 moles, zinc oxide and three moles of 02 to two moles. Zinc oxide would yield 4.5 moles of zinc oxide. The lesser amount is based on the limiting reactant. So in this case here that limiting reactant is zinc sulfide Sonal itself for the amount of oxygen required starting from the limiting reactant uh, two moles of zinc sulfide 23 moles of oxygen. And this would to yield 6.3 moles of 02 required the amount in excess. We're starting with 6.8 moles of oxygen. So that's the starting amounts. Subtract the amount that is reacted and we're left with 0.5 moles of 02 in excess

In this question, we're going to be looking at this documentary paying particular, right? More in relation to limiting and excess reach. So let's forget for example, you've got a you must be on C. And take. So what happens here is the limiting reactant is what try this chemical reaction. Oh it's what drives and determines the extent to which reform our product. Let's say for example, we've got a. That he's a lot more than big. That is we have A X. The reason why we are saying this relationship, this reaction depends on the limiting reactant. That at some point if there is a lot of gay than be at some point this reaction is going to stop because all of B is going to be consumed. But we're still going to be left with some A. Because we have said initially there is a lot of A than deep. So this going this creation is going to stop even if there is still a lot of A. That is left but it is going to stop because there is No more b. two here eight. So this entire this entire reaction was just to stop because there's no more substance of B do. It can react. So at the end of the day the formation of C and D depends on the limiting returned which is big. Now coming to what we've been. Yeah, we look at the reaction, we can tell that if we have to say for example or two against using so can tell that these are in the ratio of you're just in time. So if we've got 6.8 more or or two and we have got open to malls of open to more off Zoinks our fight, what is going to happen here is we are only going to hear 4.2 multiplied by three. So all hand 6.3 walls of oxygen needed. That is what we did here is to just look at the relationship between zinc sulfide and also So if we have got 4.2 moles of sing sulfide according to the reaction, they are in the ratio of the race. So if we've got 4.2 more things are right, The 4.2 zinc sulfide is going to need 3/2 of So 6.3 moles of oxygen are going to be needed. But we have supplied 8.6.8. So it means we've got a lot more oxygen than what we actually need. Therefore we can conclude to say question is Yeah. X. And the excess amount of focus in available is going to be 6.8 minus the oxygen that is required, which is 13 And this gives us 0.5 Yeah. These are the number of moles of excess oxygen that are in this system

We're told that we have the following reaction for every two moles of zinc sulfide that reacts with three moles of oxygen. We get two moles of zinc oxide and two moles of sulfur dioxide. Now we have a react reaction mixture that initially begins with 4.2 moles of zinc sulfide as well as 6.8 moles of Oxygen. And we want to know if this is goes to completion, how much excess reactant will remain? Well, first, we should probably figure out which reactant is excess. So let's set up now mole ratios to find out which one will produce less and more products. So according to our equation, we have two moles of zinc sulfide and let's pick a product and we'll just pick this one here for every two moles of zinc oxide. If this completed, this would be a 1 to 1 ratio and therefore we get 4.2 moles of products. Now let's check it for oxygen, which has three moles reacting, forming again two moles of zinc oxide reacting. Let's see here So 6.8 times two divided by three would be 4.53 moles of zinc oxide, which tells us that the zinc sulfide will limit the reaction of the air limiting reactant, whereas the oxygen will be our excess reactant because it makes more. But now we want to know if this went to completion and all of the zinc sulfide was used up. Since it's our limiting how much oxygen would remain, and to do that, we can do one more mole ratio. So we know we begin with 4.2 moles of zinc sulphide and according to our equation, every time this reaction proceeds, two moles of sink so fighter used up and three moles of oxygen are used up when this works out, it would indicate that 6.3 moles of oxygen are consumed every time this reaction completes. But we also know that we started with 6.8 moles. So if we subtract how much is consumed from how much it's we had, we would get that there would be 0.5 moles of oxygen presence when this reaction wraps up. Since it was excess


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