The theoretical yield of 1-bromobutane produced from 15.0 ml of butan-1-ol (density = 0.810 g/ml) is liquid and the product is a solid. Because the starting material is a liquid, you must use its density to first convert it into grams, and then you can follow the steps outlined in LSA #2 to calculate the theoretical yield. mL X g/mL = g of butan-1-ol) g. (Here the starting material is a 12.2 0.164 22.5 0.0887 13.8

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### QUESTION 20

The theoretical yield of 1-bromobutane produced from 15.0 ml of butan-1-ol (density = 0.810 g/ml) is __________ g. (Here the starting material is a liquid and the product is a solid. Because the starting material is a liquid, you must use its density to first convert it into grams, and then you can follow the steps outlined in LSA #2 to calculate the theoretical yield. *mL* x *g/mL* = *g of butan-1-ol*)

- 12.2
- 0.164
- 22.5
- 0.0887
- 13.8
Transcribed Image Text:### QUESTION 20 The theoretical yield of 1-bromobutane produced from 15.0 ml of butan-1-ol (density = 0.810 g/ml) is __________ g. (Here the starting material is a liquid and the product is a solid. Because the starting material is a liquid, you must use its density to first convert it into grams, and then you can follow the steps outlined in LSA #2 to calculate the theoretical yield. *mL* x *g/mL* = *g of butan-1-ol*) - 12.2 - 0.164 - 22.5 - 0.0887 - 13.8
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