of 15 > Tin(IV) sulfide, SnS₂, a yellow pigment, can be produced using the following reaction. SnBr(aq) + 2 Na, S(aq) →→→ 4 NaBr(aq) + SnS₂ (s) - Suppose a student adds 39.8 mL of a 0.562 M solution of SnBr4 to 50.2 mL of a 0.151 M solution of Na, S. Identify the limiting reactant. SnBr4 O Na₂S O NaBr O SnS₂ Calculate the theoretical yield of SnS₂. theoretical yield: The student recovers 0.396 g of SnS₂. Calculate the percent yield of SnS, that the student obtained.

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0 of 15 >
Tin(IV) sulfide, SnS₂, a yellow pigment, can be produced using the following reaction.
SnBr(aq) + 2 Na, S(aq) →→→ 4 NaBr(aq) + SnS₂ (s)
Suppose a student adds 39.8 mL of a 0.562 M solution of SnBr, to 50.2 mL of a 0.151 M solution of Na, S.
Identify the limiting reactant.
SnBr4
O Na₂S
NaBr
O SnS₂
Calculate the theoretical yield of SnS₂.
theoretical yield:
The student recovers 0.396 g of SnS₂. Calculate the percent yield of SnS, that the student obtained.
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Transcribed Image Text:0 of 15 > Tin(IV) sulfide, SnS₂, a yellow pigment, can be produced using the following reaction. SnBr(aq) + 2 Na, S(aq) →→→ 4 NaBr(aq) + SnS₂ (s) Suppose a student adds 39.8 mL of a 0.562 M solution of SnBr, to 50.2 mL of a 0.151 M solution of Na, S. Identify the limiting reactant. SnBr4 O Na₂S NaBr O SnS₂ Calculate the theoretical yield of SnS₂. theoretical yield: The student recovers 0.396 g of SnS₂. Calculate the percent yield of SnS, that the student obtained. E D $ 4 G Search or type URL R % 5 FL T MacBook Pro V 6 G Y B & 7 H U * 00 8 N J ( 9 K M O O < P
Expert Solution
Step 1

Given :

SnBr4 (aq) + 2 Na2S(aq) → 4 NaBr(aq) +                                                                                      SnS₂(s)

Experimental yield = 0.396 gram 

 39.8 mL of a 0.562 M solution of SnBr

50.2 mL of a 0.151 M solution of Na2

Needed information : molar mass of SnS=                                                                 182.84 g/mol 

To find : 

Identification of limiting reactant ,theoretical yield and percentage yield

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