Question

11. Sulfur hexafluoride is produced by reacting elemental sulfur with fluorine gas. Sg(s) + 24 F2(g) - 8 SF6(E) What is the percent vield im 18.3 SF is isolated from the reaction of 10.0 Ss and 300 12 a. 40.2% b. 45.89 c. 47.6% d. 54.656 e. 61.0% 12. The reaction of 5.07 g N with 0.722 g H produces 1.27NH. The percent vield of this reaction is indicate the answer choice that best completes the statement or answers the question. 73. Under certain conditions the reaction of ammonia with excess oxygen will produce a 29,596 yield of NO, Whatma of NH3 must react with excess oxygen to yield 157 g NO? 4 NH3(g) + 5 O2(g) - 4 NO(g) + 6H2O(g) a. 89.1 g b. 302 g c.263 g d. 26.3 g e. 938 g 74. Consider the fermentation reaction of glucose: C6H12062C2H5OH + 2002 A 1.00-mol sample of C6H12O6 was placed in a vat with 100 g of yeast. If 37.0 g of CzHsOH was obtained, what was the percent yield of C2H5OH? a. 40.2 % b. 20.1 % c. 37.0 % d. 100 % e, none of these 75. Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as follows: Cu2S(s) + O2(8) -2Cu(s) + SO2(8) If the reaction of 0.700 kg of copper(I) sulfide with excess oxygen produces 0.190 kg of copper metal, what is the percent yield? a. 67.9% b. 39.9 % c. 27.1% d. 135 % e. 33.9% 76. The percent yield of a chemical reaction is calculated by dividing the actual yield by the multiplying by 100%. -yield and Indicate the answer choice that best completes the statement or answers the question. 77. Nitric oxide, NO, is made from the oxidation of NH3 as follows: 4NH3 + 502-4NO + 6H2O If 9.6-g of NH3 gives 12.0 g of NO, what is the percent yield of NO? a. 93% b. 44% c. 11% d. 71% e. 28 % 78. Aspirin is produced by the reaction of salicylic acid (M = 138.1 g/mol) and acetic anhydride (M = 102.1 g/mol). CH6O3(s) + C4H603(6) - CH3O4(s) + C2H4O2(6) If 2.04 g of CH 04 (M = 180.2 g/mol) is produced from the reaction of 3.03 g CyH6O3 and 4.01 g C4H2O3, what is the percent yield? Page 10

Answer 1

Elemental Sulphur (S8) reacts with Fz (Fluorine) to produce SFG as product. The balanced chemical equation for the above reaction is, Sg +24F2 = > 8 SF6 (Imole) (24 moles) & moles) * Imole Sg reacts with 24 moles of Fą to Produce 8 móles of SF6 as product. * I mole Sz = Gram molar mass of Sg. Imole of any compound is equal to it's gram molar mass. * Molecular mass of Sg = 8x CAtomic mass of s) . = 8x(32-065) = 256.52 amu : Gram molar mass of Sg = 256.52 g 1 mole of S8 = 256.529 Molecular mass of * Similarly Molecular mass of F Fa = 2x (Atomic massa 9x = 2x(18.998) = 37.996 amu ...Gram molar mass of Fq = 37.996 g 1 mole F2 = 37.996 g * Molecular mass of SF6 = (1x(Atomic mass ofs)+6xCatomicmasoff) . -[ix(32.065)+6x (18.998)] - 32.065 +113.988] - 146.053 amu :. Grams molar mass of SF6 - 146.053 g.

... I mole of SF6 = 146.0539 * From the above, Mass of 24 moles of FQ = 24xC mass of Imole of Fa) - 24X (37.996) - 911.904 g Mass of 24 mdles * Mass of 8 miles of SF6 = 8x (Mass of 1 mole SA) = 8x (146.053) - 1168.424 g Mass of 8 moles of SF6 = 1168.4248/ Sg + 24F2 - (1more) (24 moles) . - 256529 911.9049 8 SF6. (8 moles) = 1168.4249 * According to the givers data, 30.09.06 Fal is taken for 10.09.0f Sg. From the above quantitative relation, 911-9049.0f Fz requires 256.52 g. of Sg. le 1.9048. Requines & Sg reacts with > 256.52g. of of Fg for 30.09.06 fa required Sy , ... The required Sg for 300g of Fig = 30.09 x (256.529) ( 911.9048 ) Required S8 = 8.439 g But the amount takes for Sg is 10.0g. which is more than the required (8.439 g).

:: Sp + 24 F9 8576 8.439 g 30.og only a I requires So, Fg which is taken as 30.0g is the limiting sreagent * From the quantitative Sq + 24 F 911-9049 relationship, 85F6 1168.4249 * 911.904 g. 8 Fq' produces 168.4249. of SF 91-9049. of Fa - Froduces → 1168-424 sec 30.og of Fg Produces uces ? ...Amount of SF6 Produced = 1 Bo-og) (11681424 ) L ($11.904 g) = 38-439 9. The maximum amount of SFG that can be Produced from 30.0g. of Eq = 3 (Theoretical yield) But the amount of SF6 Produced = 18. 39. . (Experimental Yield) .:: Percentage yield = Experimental yield Theoretical yield " X ( ( 18.3 g *100=47.6% (38.4399) 1. Percent yield = 47.6% Ans: [c]