ࡱ> NPMy 0bjbj .8{{0:::::NNNNT4N*j J*L*L*L*L*L*L*$5,.tp*: q\ p*:: *""" ::J*" J*""VR)@*2!N) 6**0*)x[/"[/*[/:* " p*p*" * [/  : Study Guide Even More Percent Yield Find the molar mass of the following Na3PO4 Na= 22.99x3 P= 30.97 O= 16x 4 = 163.911 g/mole MgSO4 Mg= 24.31 S= 32.07 O= 16x 4 = 120.38 g/mole Mg3(PO4)2 Mg=24.31x3 P= 30.97x2 O= 16x 8 = 262.87 g/mole Zn3(PO4)2 Zn= 65.39x3 P= 30.97x2 O= 16x 8 = 168.11 g/mole FeSO4 Fe= 55.85 S= 32.07 O= 16x 4 = 151.92 g/mole Fe3(PO4)2 Fe= 55.85x3 P= 30.97x2 O= 16x 8 = 357.49 g/mole To get the percent yield you must divide the actual yield, by the theoretical yield times 100. Can your percent yield be greater than 100%? No What is the percent yield if your actual yield is 0.00452 g and your theoretical yield was 0.184 g? .00452 g x 100 = 2.46% 0.184 g What is the percent yield if your actual yield is 93.78g and your theoretical yield was 96.8 g? 93.78 g x 100 = 96.9% 96.8 g What is the percent yield if your actual yield is 8.00g and your theoretical yield was 9.73 g? 8.00 g x 100 = 82.2% 9.73 g What is the actual yield if you know the reaction has a 16% percent yield and the theoretical yield is 64.03 g? (16%) x (64.03g) = 10 g 100 What is the actual yield if you know the reaction has a 96.8% percent yield and the theoretical yield is 0.251g? (96.8%) x (0.251g) = 0.243 g 100 What is the actual yield if you know the reaction has a 17.98% percent yield and the theoretical yield is 6.44 g? (17.98%) x (6.44g) = 50.6 g 100 Solve the following. You need to balance the equations first. Determine the percent yield for the reaction between 26.5g of MgSO4 and Na3PO4 if 18.4g of Mg3(PO4)2 is produced.  3MgSO4+ 2Na3PO4 (3Na2SO4 + Mg3(PO4)2 1) 26.5 g x 1 mole MgSO4 = 0.220136 mole MgSO4 120.38 g 2) 0.220136 mole Cs x 1 moles Mg3(PO4)2 = 0.073378 moles Mg3(PO4)2 3 mole MgSO4 3) 0.073378 moles Mg3(PO4)2 x 262.87 g = 19.29 grams Mg3(PO4)2 1 mole Mg3(PO4)2 theoretical yield 18.4 grams Mg3(PO4)2 x 100 = 95.4% 19.29grams Mg3(PO4)2 Determine the percent yield for the reaction between 28.1g of ZnCl2and excess Na3PO4 if 17.3g of NaCl is recovered along with an unknown amount of Zn3(PO4)2.  3ZnCl2+ 2Na3PO4 (6 NaCl + Zn3(PO4)2  1) 28.1 g x 1 mole ZnCl2 = 0.20618 mole ZnCl2 136.29 g 2) 0.20618 mole ZnCl2 x 6 moles NaCl = 0.4123 moles NaCl 3 mole ZnCl2 3) 0.41236 moles NaCl x 58.44 g = 24.098 grams NaCl 1 mole NaCl theoretical yield 17.3 grams NaCl x 100 = 71.79 % 24.098 grams NaCl Determine the percent yield for the reaction between 54.89g of FeSO4 and excess Na3PO4 to produce 12.8g of Fe3(PO4)2 and an unknown quantity of Na2SO4.  3FeSO4+ 2Na3PO4 (3Na2SO4 + Fe3(PO4)2  1) 54.89 g x 1 mole FeSO4 = 0.3613 moles FeSO4 132.91 g 2) 0.3613 moles FeSO4 x 1 moles Fe3(PO4)2 = 0.1204 moles Fe3(PO4)2 3 moles FeSO4 3) 0.1204 moles Fe3(PO4)2 x 357.49 g = 43.05grams Fe3(PO4)2 1 mole Fe3(PO4)2 theoretical yield 12.8 grams Fe3(PO4)2 x 100 = 29.7 % 43.05 grams Fe3(PO4)2 Determine the percent yield for the reaction between 15.8g of NaOH and excess MgSO4 to produce 9.8g of Mg(OH)2 gas and Na2SO4.  2NaOH + MgSO4 (Na2SO4 + Mg(OH)2 1) 15.8 g x 1 mole NaOH = 0.395 mole NaOH 40 g 2) 0.395 mole NaOH x 2 moles Mg(OH)2 = 0.1975 moles Mg(OH)2 2 mole NaOH 3) 0.1975 moles Mg(OH)2 x 58.33 g = 11.52 grams Mg(OH)2 1 mole Mg(OH)2 theoretical yield 9.8 grams Mg(OH)2 x 100 = 85.1 % 11.52 grams Mg(OH)2  %-KMNPQRS| * 1 2 3 7 8 9 _ f g h j k n hy2he0H*he0he0B*phhHOXhe0B*phhpphe0B* phph)he0H* h)he0 hy2he0hANB*phhHOXhANB*ph hANH*h)hANH* h)hAN hy2hANhAN hAN5\5 %&K2 3 g h   6 7 7$8$H$^gde0 & F7$8$H$gde08^8gde0gde0^gde0 & Fgde0n o p q r 3 5 7  # 7 8 % * , - 5 6 B C L T _ m n o ιwwwp he05\he0B*aJphhPuhe0B*CJaJphhpphe0B*CJaJphhe0B*CJaJphhe0>*B*CJaJphhe0B*aJphh Mhe0B*phhe0he0B*phhHOXhe0B*phhpphe0B* php hy2he0hy2he0H*+  1 8 9 % ? 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