Question

QUESTION 15 9.00 grams of C2H4 is reacted with 5.50 gram F2 according to the following equation. How many gram of HF can be formed, assuming 100% yield? (i.e., what is theoretical yield?) C2H4 + 6F2 --> 2CF4 + 4 HF Hint: Find limiting reagent. 0.281 g C 0.830 g 1.93 g 17.219 20.55 g 19.34 g 12.04 g 25.91 g QUESTION 16 When 3.50 CH4 was reacted with excess oxygen, according to the following equation, 1.45 grams of C2H2 was produced. What was the percent yield? 6 CH4 + 02 --> 2 C2H2 + 200 + 10H2 Hint: find theoretical yield Hint: percent yield actual yield/ theoretical yield)x100 13.1 % 44.2% 76.5 % OOOOOOO 63.2% 55.2% 72.2% 79.4 % 26.1 %

Answer 1

Solution question -15 Ans :- 1.939 Explanation :- C2H4 +65 + 2CF4 + 4HF here limiting suagent is E because 9 515-509) is less them.ez Hq (9.009) amount We know that, Imole F = 389F2 6 1 F2 = 228 g F2 And 1 mole HF = 209HFR av HF = 809 HF when the amount of Fz is 2289 then the amount of HF is 80 g Now, 2289 F →800HF 19 € 7 98 9HF So, 5.509 → 80x5:50 228 = 1.939 HE . Therefore the theoretical gield is 1.989

question - 16 Ans - 76.5 % Explanation : betta of O2 -> 2 C₂H₂ + 200 + 10H2 here limiting seagent is etta We know that I mole eta = 169 CHA 6 " cha = O U And, 1 mole la H₂ =269 C2H2 2 " C2H2 = 52 g Calta Νσω, 969 cha > 52geatty 4 " → 520 " 3.50 1 > 52 X 3.50 96 = 1.89.9 C2H₂ The theoretical geld is 1.89 g Now, percent yield: lactacl yield) Theoretial yield X 100 1.45 1.89 X 100 76.5%, I ; = 76.7%. ~ percent yield is 76.5%.