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Maximum mass of iodine??? Watch

1. CH3CHO + 3I2 + 4NaOH ---> CHI3 + HCOONa + 3NaI + 3H2O
The yield of triiodomethane (CHI3) obtained by a student was 83.2%

Find maximum mass of iodine that the student would have used to form 10.0g of triiodomethane. Give answer to appropriate precision.

I got 19.04g is this correct??
2. (Original post by Mariah246)
CH3CHO + 3I2 + 4NaOH ---> CHI3 + HCOONa + 3NaI + 3H2O
The yield of triiodomethane (CHI3) obtained by a student was 83.2%

Find maximum mass of iodine that the student would have used to form 10.0g of triiodomethane. Give answer to appropriate precision.

I got 19.04g is this correct??
I got 7.93g, although it is 2am, so I may well be very wrong!
3. (Original post by Mariah246)
CH3CHO + 3I2 + 4NaOH ---> CHI3 + HCOONa + 3NaI + 3H2O
The yield of triiodomethane (CHI3) obtained by a student was 83.2%

Find maximum mass of iodine that the student would have used to form 10.0g of triiodomethane. Give answer to appropriate precision.

I got 19.04g is this correct??
10g = 10/394 = 0.0254 mol

This comes from 3 x 0.0245 mol iodine = 0.0761 mol

Mass of iodine = 254 * 0.0761 = 19.3g

but his yield is only 83.2% so the actual mass he would have used is 100/83.2 * 19.3 = 23.2g
4. (Original post by charco)
10g = 10/394 = 0.0254 mol

This comes from 3 x 0.0245 mol iodine = 0.0761 mol

Mass of iodine = 254 * 0.0761 = 19.3g
but his yield is only 83.2% so the actual mass he would have used is 100/83.2 * 19.3 = 23.2g
thank you so much!!
5. (Original post by Ynang)
I got 7.93g, although it is 2am, so I may well be very wrong!
I forgot to say thanks, for trying to help
6. (Original post by charco)
10g = 10/394 = 0.0254 mol

This comes from 3 x 0.0245 mol iodine = 0.0761 mol

Mass of iodine = 254 * 0.0761 = 19.3g

but his yield is only 83.2% so the actual mass he would have used is 100/83.2 * 19.3 = 23.2g
I still don't understand the last part...

I mean, the question has told us 10.0g is formed and we've used this as the basis of our mole calculations to give the moles of I2.

Or is the question saying the yield is 82.3 % of 10 or 82.3% = 10g???
7. (Original post by ps1265A)
I still don't understand the last part...

I mean, the question has told us 10.0g is formed and we've used this as the basis of our mole calculations to give the moles of I2.

Or is the question saying the yield is 82.3 % of 10 or 82.3% = 10g???
The question is saying that the yield in the actual experiment is only 82.3%

So if you wish to make 10g of product then you actually need to aim for 10 x 100/82.3 grams of product.
8. (Original post by charco)
The question is saying that the yield in the actual experiment is only 82.3%

So if you wish to make 10g of product then you actually need to aim for 10 x 100/82.3 grams of product.
Ah thanks!!!

I also have one question about calculating moles. If I have say "2NaHCO3", do I times the Mr by 2? If not, in what cases would I times Mr by the coefficient?
9. (Original post by ps1265A)
Ah thanks!!!

I also have one question about calculating moles. If I have say "2NaHCO3", do I times the Mr by 2? If not, in what cases would I times Mr by the coefficient?
Only when you are using balanced equations to find reacting quantities ...

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