A student reacts 10.00 g of p nitrophenol (MW = 139.09 g/mol) with 6 mL of ethyl

Question

A student reacts 10.00 g of p nitrophenol (MW = 139.09 g/mol) with 6 mL of ethyl iodide (MW = 155.97 g/mol rho = 1.9500 g/mL) and suffielent sodium hydroxide to produce p - nicrophenetole (MW = 167.15 g/mol). a. What number of moles of p nicrophenol are used in the reaction? b. What mass of ethyl iodide is used in the reaction? c. What number of moles of ethyl iodide are used in the reaction? d. What is the limiting reagent in the reaction^3. e. What is the chromatic number of moles of p neurophysiology formed from the reaciton? f. If the percent yield of the reaction is 38%, what mass of p nitrophenecole was produced?

Explanation / Answer

5.

A) moles of paranitrophenol=given mass÷molar mass=10/139.09=0.07189

B) mass of ethyl iodide:- density ×volume=1.95×6=11.7 g

C) moles of ethyl iodide:-11.7÷156=0.075

D) paranitrophenol and ethyl iodide react in 1:1 ratio. So limiting reagent is paranitrophenol.

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