Question

5. Imagine that you have discovered a new way of making synthetic rubies by combining solid aluminum and oxygen gas according to the reaction below. 4 Al+3 O2 + 2 Al2O3 a. If have 10 moles of Al, and an unlimited amount of Oz, how many moles of Al2O3 could you make? b. If have 10 moles of O2, and an unlimited amount of Al, how many moles of Al2O3 could you make? c. If you were to combine 10.0 moles of Al and 10.0 moles of O2, which would be the limiting reagent? How many total moles of Al2O3 could you make? d. If you collected 2 moles of Al2O3, what is the percent yield? 6. Urea (CH4N20) is a common fertilizer that is synthesized by the reaction of ammonia (NH3) with carbon dioxide: 2 NH; (aq) + CO2 (aq) → CH4N20 (ag) + H20 (1) If an industrial synthesis of urea, a chemist combines 136.4 kg of ammonia with 211.4 kg of carbon. a. Determine the theoretical yield, in kg, of urea. b. If the chemist obtains 168.4 kg of urea, what was the percent yield of this reaction?

Answer 1

The answer is

5) YA) 4.30, –> 241,03 (a) u mole of Al produce = 2 mole A2O3 I mole Al produce = 2 mole y Al2O3 10 mole Al produce = 2x10 mole y Al2O3 - Борје Р.О. . 5 mole of Al₂O3 , can be made. ALO3 b) 3 more of Og give = a mole of I mole of a give a 2 mole of 3 ALO3 10 mole oy Og give = 2 x 10 = 20 - 6-66 more of Aboa oy Algo, will be produced :6.66 mole

- available. (C) from the ean. 3 mole o react coith - 4 more Al Imole og react with y mole Al 10 mole Og react with 4 x 10 = 40 -13.33 mole Al But only o 10 mole of Al is .Al is the limiting reagent Since 4 mole of Al give is a mole Al, Og I mole of Al give a more AlO₃ 10 mole of Al give a X10 - 5 m 1. 5 mole of Al can be made. d) Theoretical yield - 5 more Actual yield - 2 mole percentage yield - Actual yiely x100 - Theoretical - yield 2X100 = 40% in percentage yield = 40%.