Disulfur dichloride can be prepared by 3 SCl 2 + 4 NaF-------->SF 4 + S 2 Cl 2 +

Question

Disulfur dichloride can be prepared by 3 SCl2 + 4 NaF-------->SF4   +  S2Cl2+ 4NaCl What is the percent yield of the reaction if 5.00gSCl2 reacts with excess NaF to produce 1.262gS2Cl2? Explain from set up to finish, and how you did each step torecieve kharma points, Thanks for the help Disulfur dichloride can be prepared by 3 SCl2 + 4 NaF-------->SF4   +  S2Cl2+ 4NaCl What is the percent yield of the reaction if 5.00gSCl2 reacts with excess NaF to produce 1.262gS2Cl2? Explain from set up to finish, and how you did each step torecieve kharma points, Thanks for the help

Explanation / Answer

We Know that :      The given reaction is :           3SCl2 + 4 NaF--------> SF4  +  S2Cl2 + 4NaCl           number of moles of SCl2 = 5.0 g /102.971 g / mol                                                    = 0.0485 mol          from the balanced equation it is clear that 3 moles of SCl2 produces 1 mole of S2Cl2          so 0.0485mol of S2Cl2 produces ( 1 /3 x 0.0485 ) mol                                                             = 0.0161 mol             Percent yield of S2Cl2 = Theoretical yield / Actual yield * 100             Actual yield = 0.0161 mol x 135.036 g / mol                                = 2.174 g               Theoretical yield = 1.262 g ( given )                 Percentyield of S2Cl2 = 1.262 /2.174 * 100                                                      = 58.04 %

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