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http://cvms.canfield.k12.oh.us/homepages/data/canf_cc/files/ap_092313.pdf Can someone help with 3 the first one on the page with the red, blue, and purple circles?
Is the link working for you
The question that says: 2A + B --> C right? If so.. what is your interpretation of this equation?
and secondly, do you know what the limiting reagent is?
I thought it would be D because I think you have 8 of A and 5 of B. So B is your limiting.
So the most you could make is 5
And you would have 3 red/A left over?
Think about this again: in this chemical reaction you need 2 molecules of A for every molecule of B. can you see why?
Let's count them up We have 8 atoms A Red and 5 atoms of B blue right?3
the chemical reaction says that 2 molecules of A react for every molecule of B
You wouldn't multiply the 8 by 2 right and get 16?
let's go back first, we know that one of the reactants is going to run out first. 8 atoms of A and 5 atoms of B. 2A + B --> C i'm pretty sure that we need 10 atoms of A to react with 5 atoms of B. but we only have 8 atoms of A reacting with 5 atoms of B. so A will be the limiting reactant and we will be left with 1 atom of B left over BLUE.
How exactly did you figure that out?
Okay so it would be c?
I got confused about how much product we get in the end.
So looking at the reaction above, can you tell me how many molecules of each react and why? just by looking at the balanced reaction.
we will work on this step by step
2A is need for 1B to make C
so essentially this is telling us that we need twice the amount of A in order to react with B? make sense?
let's move on
So in a sense would you have to divide the A by 2?
@staldk3 we will do this step by step and you'll get it Now they gave us 8 molecules of A and 5 molecules of B.
we can do two things
First we can multiply the number of molecules by the molar ratio. do you notice that the unit we're not looking for is in the denominator and cancels out. |dw:1446862767089:dw| now we do this We make a table. We write down how many molecules we started off with for each, and how many we need if the reaction is to go to completion if there weren't any listing reagents. we then find that we need 10 molecules of A but only have 8, while we only need 4 molecules of B and have 5. That's why we have one molecule of b left over and all the A is gone. |dw:1446862891128:dw|
Rather do the mole ratio one. I've never done it the second way before I don't think.
Okay. So we would have 4 purple and 1 blue?
I think that's right.
yeah so why do you think you would get four purple that's for C the product.
Because it's limited by the limited reactant. You can only make as much as the limited reactant has.
that's correct @staldk3 but just remember yo can always check by doing this
always get into the habit of doing that whenever you have these questions multiplying by the molar ratio, ensuring that the number of moles you want for a certain reactant is in the numerator and the one you're multiplying the ratio by is in the denominator.
You always need to find out what reactant runs out FIRST that's why limiting reagents are so important
Right. Thanks for helping. I have another one like this. I'll see if I can do it. This one is a bit harder.
Make sure your equations are balanced first then. Use the molar ratios to figure how much of each reactant you need. Then make a chart like I did above and compare.
2CO + O2 -->2CO2