chem unit2 aleve;
16 Compound X forms compound Y in the reaction shown in the equation.
No knowledge of this reaction is required.
C7H6O3 + C2H4O2 → C9H8O4 + H2O
compound X compound Y
What mass of compound X is required to produce 8.4g of compound Y, if the yield is 40%?
[Molar]
A 3.4g
B 6.4g
C 16.1g
D 21.0g
why is the ans C i got it as B Thanks in advance
No knowledge of this reaction is required.
C7H6O3 + C2H4O2 → C9H8O4 + H2O
compound X compound Y
What mass of compound X is required to produce 8.4g of compound Y, if the yield is 40%?
[Molar]
A 3.4g
B 6.4g
C 16.1g
D 21.0g
why is the ans C i got it as B Thanks in advance
Original post by staerryynighht
16 Compound X forms compound Y in the reaction shown in the equation.
No knowledge of this reaction is required.
C7H6O3 + C2H4O2 → C9H8O4 + H2O
compound X compound Y
What mass of compound X is required to produce 8.4g of compound Y, if the yield is 40%?
[Molar]
A 3.4g
B 6.4g
C 16.1g
D 21.0g
why is the ans C i got it as B Thanks in advance
No knowledge of this reaction is required.
C7H6O3 + C2H4O2 → C9H8O4 + H2O
compound X compound Y
What mass of compound X is required to produce 8.4g of compound Y, if the yield is 40%?
[Molar]
A 3.4g
B 6.4g
C 16.1g
D 21.0g
why is the ans C i got it as B Thanks in advance
Can you please show your working? So we can tell you where you went wrong.
Original post by Pigster
Can you please show your working? So we can tell you where you went wrong.
okay here is my working
Original post by staerryynighht
okay here is my working
You ignored the "if the yield is 40%" bit.
Original post by Pigster
You ignored the "if the yield is 40%" bit.
even when i multiply with 40% i get 2.567 which isnt any of the answers
Original post by staerryynighht
even when i multiply with 40% i get 2.567 which isnt any of the answers
Well you can't just ignore it.
Write down the equation for percentage yield and see how you could apply it to this situation. You'll no doubt have to de a little algebraic rearranging.
so i put actual mass over theoritical timess 100 ugh i still cant seem to reach the ans pls help
Original post by staerryynighht
16 Compound X forms compound Y in the reaction shown in the equation.
No knowledge of this reaction is required.
C7H6O3 + C2H4O2 → C9H8O4 + H2O
compound X compound Y
What mass of compound X is required to produce 8.4g of compound Y, if the yield is 40%?
[Molar]
A 3.4g
B 6.4g
C 16.1g
D 21.0g
why is the ans C i got it as B Thanks in advance
No knowledge of this reaction is required.
C7H6O3 + C2H4O2 → C9H8O4 + H2O
compound X compound Y
What mass of compound X is required to produce 8.4g of compound Y, if the yield is 40%?
[Molar]
A 3.4g
B 6.4g
C 16.1g
D 21.0g
why is the ans C i got it as B Thanks in advance
Original post by staerryynighht
even when i multiply with 40% i get 2.567 which isnt any of the answers
you get 0.1167 when u do 40%
and then you times that by the mr 138 to get 16.1
credits to @trusfratedblink for working it out haha
(edited 3 years ago)
Original post by staerryynighht
so i put actual mass over theoritical timess 100 ugh i still cant seem to reach the ans pls help
act/theory x 100 = yield
You know your act and your yield, so you must have to work out the theory mass (of product)... so what mass of reactant is needed to make that theory mass?
Original post by squishyraisin
you get 0.1167 when u do 40%
and then you times that by the mr 138 to get 16.1
credits to @trusfratedblink for working it out haha
and then you times that by the mr 138 to get 16.1
credits to @trusfratedblink for working it out haha
i cant seem to get 0.1167 please help
Original post by staerryynighht
i cant seem to get 0.1167 please help
You literally divide by 0.4 because it’s 40%
Original post by staerryynighht
i cant seem to get 0.1167 please help
Original post by squishyraisin
you get 0.1167 when u do 40%
and then you times that by the mr 138 to get 16.1
credits to @trusfratedblink for working it out haha
and then you times that by the mr 138 to get 16.1
credits to @trusfratedblink for working it out haha
I've attached my working out, I hope this helps
Original post by trusfratedblink
I've attached my working out, I hope this helps
aww tysm do u mind if i ask another q
Original post by staerryynighht
aww tysm do u mind if i ask another q
sure, I'll try to be as helpful as I can
Original post by trusfratedblink
sure, I'll try to be as helpful as I can
here i tried doing this but got stuck
Original post by staerryynighht
here i tried doing this but got stuck
Like I said in your other thread: you haven't balanced the equation.
Original post by Pigster
Like I said in your other thread: you haven't balanced the equation.
i balanced it now and took the mole ratio 1:1 for bariumhydroxide ND BARIUM NITRATE then i wanted to find the concentration of barium nitrate and the ans is 2.0 but the ms says B
Original post by staerryynighht
i balanced it now and took the mole ratio 1:1 for bariumhydroxide ND BARIUM NITRATE then i wanted to find the concentration of barium nitrate and the ans is 2.0 but the ms says B
There are two ways of doing it.
1. work out the number of mol of everything. -> 0.05 mol of Ba(OH)2 and 0.1 mol of HNO3. You recognise from the balanced equation that 1 mol of Ba(OH)2 reacts wtih 2 mol of HNO3, hence there is the right amount of both. From the equation you realise that 0.05 mol of Ba(NO3)2 forms. Then to work out the concentration you divide amount by volume. Since you've added 50 cm3 to 50 cm3, the volume is now 100 cm3 and hence conc. is 0.50 mol dm-3
2. the Q states "Exactly" so you don't need to bother doing all the mol bit. In the back of you mind you know that the formulae are Ba(OH)2 and Ba(NO3)2 so you know that it must be a 1:1 ratio as both contains only 1 Ba2+ ion. There must be a the same number of mol of both. But you notice that the volume has doubled, hence the conc. must have halved. Since the conc. initially was 1.00 it must now be B.
Original post by staerryynighht
here i tried doing this but got stuck
here you go, I attached my working out, included the steps for solving this question. Hope it helps!
Apologies for my hand's shadow!Quick Reply
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