I need help with number one I wrote the moles for the first two parts. After com

Question

I need help with number one I wrote the moles for the first two parts. After completed the experiment I put how much t butyl chloride which was 8.2 mL. Thanks 9-t-BUTYL CHLORIDE Name Grades: Product Report Ch 2.57 1. Supply the requested information. a. Moles of t-butyl alcohol introduced. VDA owcwm b. Moles of HCL introduced. m. 5lo. Ilo21 Cwcom c. The limiting reagent is d. Theoretical yield-moles-of t-butyl chloride. e. Theoretical yield-grams-of t-butyl chloride. Actual yield grams. g. Percent yield of t-butyl chloride. 2. Why was the excess hydrochloric acid neutralized by the additon of 5% aqueous sodium bicarbonate rather than an aqueous solution of sodium hydroxide? ed for t

Explanation / Answer

1 mole of tert-butyl alcohol = 74.12 9

for 10.9 g = 10.9/74.12 = 0.14 moles

1 mole of hydrochloride = 36.45 g

for 50 ml

mass = density x volume = 1.49 x 50 = 74.5 g

for 74.5 g = 74.5/36.45 = 2.04 moles

therefore the limiting reagent is tert-butyl alcohol

74.12 g tert-butyl alcohol will produce 92.57 g of tert-butyl chloride which is 100 % yield

then for 10.9 g = 10.9 x 92.57/74.12 = 13.61 g is 100 % yield

actual yield of tert butyl chloride in grams = Density x volume = 8 x .840 g = 6.88 g

therefore the actual yield = 6.88/13.61 x 100 = 50 %

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