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Stuck on Chemistry Equation

The question is as follows:

Calculate the mass of sodium chloride that can be made from one mole of sodium in the reaction: 2Na+CL(2)----> 2NaCl

So basically I changed the equation to this:

Na+Cl(2)----> NaCl(2)
But I found my answer was wrong and only proved to be correct when I changed the equation to this:

Na+Cl----> NaCl

Why do you have to change the amount of chlorine? Is it because I basically have to change the entire proportion of the reactants by changing one to make it reactable?

Thanks in advance :>
Reply 1
Avatar for forsaker
forsaker
3
mass of Na = moles x MR
it says one mole of sodium in the question so 1 x 35.5 = 35.5g

the molar ratio between Na:NaCl is 2:2 (from the equation)
2:2 is the same as 1:1
so one mole of NaCl.

mass of NaCl = moles x Mr = 1 x (23+35.5)= 58.5g

your equation worked because the molar ratio in the original equation (2:2) is the same as the one in your second dumbass'd equation (1:1)
Reply 2
Avatar for abbeyb17
abbeyb17
I've been practicing this for my module in Jan, so hopefully this will be alright:

Essentially, you don't/shouldn't change the equation at all. You should work through the respective Relative Molecular Masses, and get yourself the answer - put things in brackets as you work it out and just get rid of them as you work out the contents - so for that you'd do the masses, and divide by 2 to gain the answer for one mole. Does that make any sense?
Original post by forsaker

Original post by forsaker
mass of Na = moles x MR
it says one mole of sodium in the question so 1 x 35.5 = 35.5g

the molar ratio between Na:NaCl is 2:2 (from the equation)
2:2 is the same as 1:1
so one mole of NaCl.

mass of NaCl = moles x Mr = 1 x (23+35.5)= 58.5g

your equation worked because the molar ratio in the original equation (2:2) is the same as the one in your second dumbass'd equation (1:1)


Thanks
Original post by abbeyb17

Original post by abbeyb17
I've been practicing this for my module in Jan, so hopefully this will be alright:

Essentially, you don't/shouldn't change the equation at all. You should work through the respective Relative Molecular Masses, and get yourself the answer - put things in brackets as you work it out and just get rid of them as you work out the contents - so for that you'd do the masses, and divide by 2 to gain the answer for one mole. Does that make any sense?



Ahh ok.
Reply 5
Avatar for user152713
user152713
Original post by abbeyb17
I've been practicing this for my module in Jan, so hopefully this will be alright:

Essentially, you don't/shouldn't change the equation at all. You should work through the respective Relative Molecular Masses, and get yourself the answer - put things in brackets as you work it out and just get rid of them as you work out the contents - so for that you'd do the masses, and divide by 2 to gain the answer for one mole. Does that make any sense?


AS OCR right?
Original post by forsaker

Original post by forsaker
mass of Na = moles x MR
it says one mole of sodium in the question so 1 x 35.5 = 35.5g

the molar ratio between Na:NaCl is 2:2 (from the equation)
2:2 is the same as 1:1
so one mole of NaCl.

mass of NaCl = moles x Mr = 1 x (23+35.5)= 58.5g

your equation worked because the molar ratio in the original equation (2:2) is the same as the one in your second dumbass'd equation (1:1)


I'm also stuck on this question. Please help :<
5.2g of Potassium chloride, KCL, was made from 5.6g of potassium hydroxide, KOH. What was the percentage yield?
Original post by Believeandsucceed
I'm also stuck on this question. Please help :<
5.2g of Potassium chloride, KCL, was made from 5.6g of potassium hydroxide, KOH. What was the percentage yield?


I think it's 5.2/5.6= 92%
Reply 8
Avatar for nadiaha
nadiaha
Your not really changing the amounts of any of the substances you are just simplifying the balanced equation going from a ratio of 2:2 to 1:1 which makes it a bit less complicated.

For that second question. percentage yield of which chemical?
Original post by Believeandsucceed
I'm also stuck on this question. Please help :<
5.2g of Potassium chloride, KCL, was made from 5.6g of potassium hydroxide, KOH. What was the percentage yield?


Percentage yield=Actual yieldTheoretical yield×100Percentage\ yield = \frac{Actual\ yield}{Theoretical\ yield} \times 100

Since the law of conservation of mass says that the mass of reactants = the mass of the products, the theoretical yield is 5.6g. We already know that the actual yield is 5.2g so just put them into the equation. 5.2 divided by 5.6 gives you 0.92, times 100 gives you 92%.
Reply 10
Avatar for Maker
Maker
20
5.6g of KOH is 0.1 mole of KOH

It should yield 0.1 mol of KCl which is 7.45g if it was 100% conversion

but it yielded 5.2 of KCl which is 5.2/7.45 = 0.7

The yield is 70%
(edited 13 years ago)
Original post by iPthreefifthteen

Original post by iPthreefifthteen
I think it's 5.2/5.6= 92%


That's what I thought but the answers at the back of the book said: 69.8% So confused!
Original post by ThisIsOurDecision

Original post by ThisIsOurDecision
Percentage yield=Actual yieldTheoretical yield×100Percentage\ yield = \frac{Actual\ yield}{Theoretical\ yield} \times 100

Since the law of conservation of mass says that the mass of reactants = the mass of the products, the theoretical yield is 5.6g. We already know that the actual yield is 5.2g so just put them into the equation. 5.2 divided by 5.6 gives you 0.92, times 100 gives you 92%.


Nope the answers say its 69.8%
Original post by Believeandsucceed
That's what I thought but the answers at the back of the book said: 69.8% So confused!


Damn :frown: chemistry is a pain in the a**
Original post by Believeandsucceed
Nope the answers say its 69.8%


Just realised that I've done the atom economy of the reaction (Mass desired product over total mass of reactants) instead of the yield. The guy above has it right.
Reply 15
Avatar for abbeyb17
abbeyb17
Original post by Rassam
AS OCR right?


Nope, GCSE AQA C2, although we've started C3 and I'm not sure where we started each module so I'm just revising everything we did since our last module.
Reply 16
Avatar for yoyo1994
yoyo1994
im also doin the c2 exam in jan 2011. its quite difficuly. u need to know everything in so much detail.

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