1) The theoretical yield of SnS2 will be 4.20 grams
2) The percent yield will be 7.93%
From the equation of the reaction:
[tex]SnBr_4(aq)+2Na_2S(aq)-- > 4NaBr(aq)+SnS_2(s)[/tex]
The mole ratio of the reactant is 1:2.
Mole of 48.1 mL, 0.478 M SnBr4 = 0.478 x 48.1/100 = 0.023 mols
Mole of 48.8 mL, 0.160 M Na2S = 0.160 x 48.8/1000 = 0.0078 moles
Thus, Na2S is in excess while SnBr4 is limiting.
Mole ratio of SnBr4 and SnS2 = 1:1
Equivalent mole of SnS2 = 0.023 moles
Mass of 0.023 noles SnS2 = 0.023 x 182.81 = 4.20 grams
2) With 0.0333 g of SnS2 recovered, percent yield = 0.333/4.2 x 100 = 7.93%
More on percent yield can be found here: https://brainly.com/question/17042787
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Here is the complete question:
Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.
SnBr4(aq)+2Na2S(aq)⟶4NaBr(aq)+SnS2(s)
Suppose a student adds 48.1 mL of a 0.478 M solution of SnBr4 to 48.8 mL of a 0.160 M solution of Na2S.
1) Calculate the theoretical yield of SnS2. ;
2) The student recovers 0.333 g of SnS2. Calculate the percent yield of SnS2 that the student obtained.
