How many moles of NaI are needed to react with 0.10 mL of 1-chlorobutane? How many moles of NaI are needed to react with 0.10 mL of 1-bromobutane?

What do you think? Please show some attempt to this question so we can better help you.

C4H9Cl + NaI -------C4H9I + NaCl

 

Answer: 1 mole

 

and ditto for the second reaction

Two questions for you to think about/investigate.

 

1) Will the iodide replace/be replaced by both the chloride and the bromide?

 

2) How much of each reagent are you starting with?

 

 

 

0.10 mL

1.Yes

2. 0.10 ml

 

0.10 moles?

Actually, the problem is written just like that with no starting amounts or molarity, which is confusing me. I do see that in the Materials sections it lists: 18% sodium iodide in acetone.

How many moles of NaI are needed to react with 0.10 mL of 1-chlorobutane? How many moles of NaI are needed to react with 0.10 mL of 1-bromobutane?

 

Are chlorobutane and bromobutane liquids?

If so can you not work out how many moles there are in 0.1 mL of pure substance.

It will be different in each case.

 

http://www.odinity.com/nucleophilic-substitution/

2ml of 18% sodium iodide are react with 0.1ml of 1-chlorobutane. Then 2ml of 18% sodium iodide react with 0.1ml of 1-bromobutane.

 

2ml x18grams/100ml x 1mole/150g= 0.0024 mole NaI react with 1-chlorobutane and 1-bromobutane, respectively. (conc. are not included for chlorobutane /bromobutane)

 

Is 0.0024 mole NaI correct?

You don't need concentration. It is assumed based on the question that the compounds are liquids. Are you familiar with density?

0.1mL x 0.89gm/ml x1mole/92.57g chlorobutane x1moleNaI/1moleChlorobutane=9.614x10^-4 NaI mole

 

0.1mL x1.27gm/ml x 1mole/137.02g bromobutane x 1moleNaI/Bromobutane=9.268 x 10^-4 NaI mole

 

How's that? I see that this is a limiting reagent problem.

 

 

But don't you have to multiply by 0.1mL the volume of the reagent?

 

How many moles of NaI are need to react with 0.10mL of chlorobutane/bromobutane? Hence 10^-4 power

Edited March 16, 20178 yr by gammagirl

0.1mL x 0.89gm/ml x1mole/92.57g chlorobutane x1moleNaI/1moleChlorobutane=9.164x10^-4 NaI mole

 

0.1mL x1.27gm/ml x 1mole/137.02g bromobutane x 1moleNaI/Bromobutane=9.268 x 10^-4 NaI mole

 

How's that? I see that this is a limiting reagent problem.

 

Well you are getting there but your arithmetic is suspect.

 

I make the chlorobutane 0.89/92.6 = .0096 moles or 9.6 x 10^-3 moles.

 

and the bromobutane 1.27/137 = .00927moles or 9.3 x 10^-3 moles.

 

Edit also present your calculation as a division not a multiplication please.

Edited March 15, 20178 yr by studiot

0.1mL x 0.89gm/ml x1mole/92.57g chlorobutane x1moleNaI/1moleChlorobutane=9.614x10^-4 NaI mole

 

0.1mL x 0.89gm/ml x1mole/92.57g chlorobutane x1moleNaI/1moleChlorobutane=9.614x10^-4 NaI mole

 

0.1mL x1.27gm/ml x 1mole/137.02g bromobutane x 1moleNaI/Bromobutane=9.268 x 10^-4 NaI mole

 

How's that? I see that this is a limiting reagent problem.

 

 

But don't you have to multiply by 0.1mL the volume of the reagent?

 

How many moles of NaI are need to react with 0.10mL of chlorobutane/bromobutane? Hence 10^-4 power

 

0.1mL x1.27gm/ml x 1mole/137.02g bromobutane x 1moleNaI/Bromobutane=9.268 x 10^-4 NaI mole

 

How's that? I see that this is a limiting reagent problem.

 

hypervalent_iodine,

Can you check the above problems?

 

Here is one more : How many moles of NAI if using 2ml of 18% sodium Iodide?

My answer: 2mL x 18grams/100mL x 1mole/150gram = 0.0024 mole

 

 

But don't you have to multiply by 0.1mL the volume of the reagent?

 

How many moles of NaI are need to react with 0.10mL of chlorobutane/bromobutane? Hence 10^-4 power

 

Well you are getting there but your arithmetic is suspect.

 

I make the chlorobutane 0.89/92.6 = .0096 moles or 9.6 x 10^-3 moles.

 

and the bromobutane 1.27/137 = .00927moles or 9.3 x 10^-3 moles.

 

Edit also present your calculation as a division not a multiplication please.

You would be correct if the question stated 1 mole, but it didn't. It asked for how much NaI is required when 0.1 mL reacts of each liquid, hence the power should be to the -4.

 

0.1mL x 0.89gm/ml x1mole/92.57g chlorobutane x1moleNaI/1moleChlorobutane=9.614x10^-4 NaI mole

 

0.1mL x 0.89gm/ml x1mole/92.57g chlorobutane x1moleNaI/1moleChlorobutane=9.614x10^-4 NaI mole

 

0.1mL x1.27gm/ml x 1mole/137.02g bromobutane x 1moleNaI/Bromobutane=9.268 x 10^-4 NaI mole

 

How's that? I see that this is a limiting reagent problem.

 

 

But don't you have to multiply by 0.1mL the volume of the reagent?

 

How many moles of NaI are need to react with 0.10mL of chlorobutane/bromobutane? Hence 10^-4 power

0.1mL x1.27gm/ml x 1mole/137.02g bromobutane x 1moleNaI/Bromobutane=9.268 x 10^-4 NaI mole

 

How's that? I see that this is a limiting reagent problem.

 

hypervalent_iodine,

Can you check the above problems?

 

Here is one more : How many moles of NAI if using 2ml of 18% sodium Iodide?

My answer: 2mL x 18grams/100mL x 1mole/150gram = 0.0024 mole

 

 

But don't you have to multiply by 0.1mL the volume of the reagent?

 

How many moles of NaI are need to react with 0.10mL of chlorobutane/bromobutane? Hence 10^-4 power

You are correct. StudioT didn't account for the correct volume.

 

Will have to check your other question in a bit.

Edit: I get the same answer as you do for your other question as well. Just be careful of the molar masses you use. Rounding errors carry through when you do larger calculations, and can make your final answer out by enough for you to lose marks.

I appreciate you, thank-you.

How many moles of NaI are needed to react with 0.10 mL of 1-chlorobutane? How many moles of NaI are needed to react with 0.10 mL of 1-bromobutane?

step 1) ALWAYS write the equation.

Step 2) work out the amount of moles of the product you are given information about

Step 3) use the equation to find ratio

step 4) find the moles of NaI by multiplying the ratio of NaI with step 2 answer but only if NaI has a bigger proportion in step 3 e.g. 3:1 (otherwise divide it)

 

HypervalentIodine

You would be correct if the question stated 1 mole, but it didn't. It asked for how much NaI is required when 0.1 mL reacts of each liquid, hence the power should be to the -4.

 

Well spotted, thanks, +1.

That was an embarrassingly silly slip.

Thank-you both.