Question

The reaction between dihydrogen sulfide and sulfur dioxide is outlined below. 2 H2S(g) S02(g) -Y 3 S(s) 2 H20(g) a. Identify the limiting reactant when 3.89 g of dihydrogen sulfide react with 4.11 g of sulfur dioxide. Justify your answer. d. Based on your answer from part (a), determine the maximum mass of sulfur that can be produced in this reaction. c. Ifthe actual yield of sulfur is found to be 4.89 g, find the percent yield in this reaction.

Answer 1

Answer:

Explanation:

2 H₂S(g) +S0₂(g) =  3 S(s) +  2H₂0(g)

2 x 34 g     64 g        3 x 32 g

68 g of  H₂S reacts with 64 g of S0₂

3.89 g of H₂S reacts with 64 x 3 .89 / 68 g of S0₂

3.89 g of H₂S reacts with 3.66  g of S0₂

S0₂ given is 4.11 g , so it is in excess .

Hence H₂S is limiting reagent .

68 g of  H₂S reacts with  S0₂ to give 96 g of Sulphur

3.89 g of  H₂S reacts with  S0₂ to give 96 x 3.89 / 68 g of Sulphur

3.89 g of  H₂S reacts with  S0₂ to give 96 x 3.89 / 68 g of Sulphur

5.49 g of Sulphur is produced .

Actual yield is 4.89

percentage yield = 4.89 x 100 / 5.49

= 89 % .