Question

Tin(IV) sulfide, Sns,, a yellow pigment, can be produced using the following reaction. SnBr.(aq) + 2 Na, S(aq) - 4 NaBr(aq) + SnS,(s) Suppose a student adds 40.0 mL of a 0.492 M solution of SnBr, to 42.6 mL of a 0.127 M solution of Na, S. Identify the limiting reactant. NaBr Na s Osns SnBr. Calculate the theoretical yield of SnSz. theoretical yield: The student recovers 0.313 g of Sns. Calculate the percent yield of Sns, that the student obtained. percent yield:

Answer 1

[sn Bru] = 0.492M, come nsnibru = 10.492x40) = 19.68 mmol In Bry (99) + 2NC2S (ez initial 19.68 5.4102 (mmol) (Na2S] = 0.1271, 42.6m1 in Noys = (0.127 x 42.6) = 5.4102 mmol u Nabr (9) + SnS2 (s) Cimiting 19.08 5.4102 Reagent 2 2 s so, was will act as limiting Reagent Limiting Reagent = Nas an > nNags = nsns2 => nonsz= 5.4102 mmol MMsns, = 118.7 + 2132.0G) => 182.82 gmimol Wonsz = 5.4 102 x 103 x 182.82 theo yield W sns₂ = 0.494 gm any actucel yield = 0.313 gm % yield = actucel Yield xico :) 0:313 xico theo yield 2 0.494 1% yield = 63.30% ) dry