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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.

Respuesta :

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 Hâ‚‚S reacts with 64 x 3 .89 / 68 g of S0â‚‚

3.89 g of H₂S reacts with 3.66  g of S0₂

S0â‚‚ given is 4.11 g , so it is in excess .

Hence Hâ‚‚S 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 % .  

CintiaSalazar

Considering the reaction stoichiometry and the definition of percent yield:

a. Hâ‚‚S will be the limiting reagent.

b. the maximum mass of sulfur that can be produced in the reaction is 5.49 grams.

c. the percent yield of the reaction is 89%.

The balanced reaction is:

2 H₂S(g) + SO₂(g) → 3 S(s) +  2 H₂O(g)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Hâ‚‚S: 2 moles
  • SOâ‚‚: 1  mole
  • S: 3 moles  
  • Hâ‚‚O: 2 moles

The molar mass of each compound is:

  • Hâ‚‚S: 34 g/mole
  • SOâ‚‚: 64 g/mole
  • S: 32 g/mole  
  • Hâ‚‚O: 18 g/mole

So, by reaction stoichiometry, the following amounts of mass of each compound participate in the reaction:

  • Hâ‚‚S: 2 molesĂ— 34 g/mole= 68 grams
  • SOâ‚‚: 1 molesĂ— 64 g/mole= 64 grams
  • S: 3 molesĂ— 32 g/mole= 96 grams
  • Hâ‚‚O: 2 molesĂ— 18 g/mole= 36 grams

a. Limiting reagent

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 64 grams of SOâ‚‚ reacts with 68 grams of Hâ‚‚S, 4.11 grams of SOâ‚‚ react with how much mass of Hâ‚‚S?

[tex]mass of H_{2} S=\frac{4.11 grams of SO_{2}x 68 grams of H_{2} S }{64 grams of SO_{2}}[/tex]

mass of Hâ‚‚S= 4.37 grams

But 4.37 moles of Hâ‚‚S are not available, 3.89 grams are available. Since you have less mass than you need to react with 4.11 grams of SOâ‚‚, Hâ‚‚S will be the limiting reagent.

b. Maximum mass of sulfur

Using the limiting reagent, you can apply the following rule of three: if by stoichiometry 68 grams of Hâ‚‚S produce 96 grams of S, 3.89 grams of Hâ‚‚S will produce how much mass of S?

[tex]mass of S=\frac{96 grams of Sx 3.89 grams of H_{2} S }{68 grams of H_{2}S}[/tex]

mass of S= 5.49 grams

So, the maximum mass of sulfur that can be produced in the reaction is 5.49 grams.

c. Percent yield

The amount of product that is obtained when all the limiting reagent reacts.  This is called the theoretical yield of the reaction. That is, the theoretical yield is the maximum amount of product that can be produced in a reaction.

On the other hand, the actual yield is the amount of product actually obtained from a reaction.

The percent yield determines the efficiency of the  reaction, and describes the ratio of the actual yield to the theoretical yield:

[tex]percent yield=\frac{actual yield}{theoretical yield}x100[/tex]

In this case, you know:

  • actual yield= 4.89 g
  • theoretical yield= 5.49 g

So, the percent yield can be calculated as:

[tex]percent yield=\frac{4.89 grams}{5.49 grams}x100[/tex]

Solving:

percent yield= 89%

Finally, the percent yield of the reaction is 89%.

Learn more about reaction stoichiometry:

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