The theoretical yield of [tex]SnS_2[/tex] will be 4.20 grams while the percent yield will be 7.93%
From the equation of the reaction, the mole ratio of [tex]SnBr_4[/tex] to [tex]Na_2S[/tex] is 1:2.
Mole of 48.1 mL, 0.478 M [tex]SnBr_4[/tex] = 0.478 x 48.1/100 = 0.023 mols
Mole of 48.8 mL, 0.160 M [tex]Na_2S[/tex] = 0.160 x 48.8/1000 = 0.0078 moles
[tex]SnBr_4[/tex][tex]Na_2S[/tex] is the limiting reactant.
Mole ratio of [tex]SnBr_4[/tex] and [tex]SnS_2[/tex] = 1:1
Equivalent mole of [tex]SnS_2[/tex] = 0.023 moles
Mass of 0.023 noles [tex]SnS_2[/tex]= 0.023 x 182.81 = 4.20 grams
With 0.0333 g of [tex]SnS_2[/tex] recovered, percent yield = 0.333/4.2 x 100 = 7.93%
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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.
