gammagirl Posted March 15, 2017 Share Posted March 15, 2017 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? Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted March 15, 2017 Share Posted March 15, 2017 What do you think? Please show some attempt to this question so we can better help you. Link to comment Share on other sites More sharing options...
gammagirl Posted March 15, 2017 Author Share Posted March 15, 2017 C4H9Cl + NaI -------C4H9I + NaCl Answer: 1 mole and ditto for the second reaction Link to comment Share on other sites More sharing options...
studiot Posted March 15, 2017 Share Posted March 15, 2017 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 Link to comment Share on other sites More sharing options...
gammagirl Posted March 15, 2017 Author Share Posted March 15, 2017 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. Link to comment Share on other sites More sharing options...
studiot Posted March 15, 2017 Share Posted March 15, 2017 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/ Link to comment Share on other sites More sharing options...
gammagirl Posted March 15, 2017 Author Share Posted March 15, 2017 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? Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted March 15, 2017 Share Posted March 15, 2017 You don't need concentration. It is assumed based on the question that the compounds are liquids. Are you familiar with density? 1 Link to comment Share on other sites More sharing options...
gammagirl Posted March 15, 2017 Author Share Posted March 15, 2017 (edited) 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, 2017 by gammagirl Link to comment Share on other sites More sharing options...
studiot Posted March 15, 2017 Share Posted March 15, 2017 (edited) 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, 2017 by studiot 2 Link to comment Share on other sites More sharing options...
gammagirl Posted March 22, 2017 Author Share Posted March 22, 2017 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 Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted March 22, 2017 Share Posted March 22, 2017 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. 2 Link to comment Share on other sites More sharing options...
gammagirl Posted March 22, 2017 Author Share Posted March 22, 2017 I appreciate you, thank-you. 1 Link to comment Share on other sites More sharing options...
Dr.Science Posted March 29, 2017 Share Posted March 29, 2017 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) Link to comment Share on other sites More sharing options...
studiot Posted March 29, 2017 Share Posted March 29, 2017 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. Link to comment Share on other sites More sharing options...
gammagirl Posted March 29, 2017 Author Share Posted March 29, 2017 Thank-you both. Link to comment Share on other sites More sharing options...
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