Question

Please help! Consider the following reaction: 3H2S (g) + 3O2 (g)  2 SO2 (g) + 2H2O (g) If O2 was the excess reagent, 4.15 mol of H2S were consumed, and 68.55 g of water were collected after the reaction has gone to completion, what is the percent yield of the reaction? Show your work.

Answer 1

Answer:

91.8 %

Explanation:

Data given:

Amount of H₂S = 4.15 mol

Amount of water (H₂O) = 68.55 g

Amount of oxygen O₂ = in excess

Percent yield of reaction water (H₂O) of water = ?

Reaction Given:

          2H₂S (g) + 3O₂ (g) ----------> 2SO₂ (g) + 2H₂O (g)

Solution:

First we look for the theoretical yield by looking in the reaction

          2H₂S (g) + 3O₂ (g) ----------> 2SO₂ (g) + 2H₂O (g)

As oxygen is in the excess so only H₂S amount act as limiting reagent.

           2H₂S (g) + 3O₂ (g) ----------> 2SO₂ (g) + 2H₂O (g)

             2 mol                                                       2 mol

to convert amount of H₂O from moles to grams

           mass in grams = no. of moles x molar mass

Molar Mass of H₂O = 2(1) + 16 = 18 g/mol

Put values in above equation

          mass in grams = 2 mol x 18 g/mol

          mass in grams = 36 g

So,

             3H₂S (g) + 3O₂ (g) ----------> 2SO₂ (g) + 2H₂O (g)

             2 mol                                                         36 g

As from the reaction it is clear that 2 mole of H₂S gives 36g H₂O then 4.15 mole will give how many grams of water

Apply unity formula

                         2 mol of H₂S ≅ 36 g of H₂O

                         4.15 mol of H₂S ≅ X g of H₂O

Do cross multiplication

                X g of H₂O =  36 g x 4.15 mol / 2 mol

                X g of H₂O =  74.7 g

So theoretical yield =  74.7 g of H₂O

Formula used for percent yield

            percent yield = actual yield / theoretical yield x 100

Put values in above equation

           percent yield = 68.55 g / 74.7 g x 100

           percent yield = 91.8 %

***Note

For SO₂ it is important to have actual yield. and implement same work.