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.
