Please answer all of them ! HAC Na-Br Br Na-l 40 mmol mw 149.89 mw 150 mw 137 mw

Question

Please answer all of them ! HAC Na-Br Br Na-l 40 mmol mw 149.89 mw 150 mw 137 mw 184. d 31.27 g/ml d 1.62 g/ml How many milliliters of 1-bromobutane should the chemist add to the flask? How many grams of Nal should the chemist use? If the chemist obtained 5.52 grams product at the end of the reaction, what was the percent yield? How many grams of the product, 1-iodobutane could theoretically be produced? If the 5.52 grams 1-iodobutane were purified and 2.0 mL were obtained, what was the recovery

Explanation / Answer

a)

mL of 1-bromobutane required

mass = mol*MW = (40*10^-3)(137) = 5.48 g

V = m/D = 5.48/(1.27) = 4.315 mL required

b)

Mass of NaI

since ratio is 1:1, we require 40 mmol of NaI

mass = mol*MW = (40*10^-3)(149.89) = 5.9956 g of NaI

c)

m = 5.52 g of product... find %yield

mas of 40 mmol = 184*40*10^-3 = 7.36 g of product expected

%yield = real/theoretical * 100 = 5.52/7.36*100 = 75%

d)

mass of produc should be produced

from above, 7.36 g of product should be produced

e)

m = 5.52 g of Iodobutane and V = 2mL

find recoery

volume = M/D = 5.52/1.62 = 3.407 mL

Recovery = recoevered/total * 100 = 2/3.407*100 = 58.7026%

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