Question

Heted 15 out of 20 Suppose a student als 42.3 mL of a 0.412 M solution of SnBr, to 41.3 mL of a 0.121 M solution of Na,s. Identify the limiting reactant. Na,s NaBr SnBr, Sns, Calculate the theoretical yield of Sus, theoretical yield: The student recovers 0.243 g of Sns, Calculate the percent yield of Sns, that the student obtained. percent yield:

Answer 1

+ 4 NaBr SnBr4 +2Na₂S - SnS2 No. of moles = Molarity & Volume (mi) 1000 No of moles of Su B84 = 0.412 x 42.3 - = 0.0174276 1000 No. of moles of Na₂S = 0.121 X 41.3 = 0. 49973 Tooo (1) - From the equation 1 mole Su Bry reacts with 2 mole Nags. 0.0174276 - 2 (0.0174276) = 0.0348552 but there is no 0.0348552 amat of Nags. . Nags is limiting agent Calculation of theoretical field of Sus2 from the equation 2 moles of Nags 1 mole of SuS₂ : 0.0049973 moles = 0.0049973 x = 0.00249865 moles (11) 2 mwt of sus2 = 118.71+ 2(32) = 182.71 amount (mass) of sus,= moles & mot = 0.00249865 x 182.71 = 0.4565g. (iii) Calculation of % Yield of suse amount obtained X100= ſo 0.243 x100 yield Theoriticlamont o.u565 = 53.23%