In: Chemistry
Elemental S reacts with O2 to form SO3 according to the reaction
2S+3O2→2SO3
A) How many O2 molecules are needed to react with 4.13 g of S?
Express your answer numerically in units of molecules.
B) What is the theoretical yield of SO3 produced by the quantities described in Part A?
Express your answer numerically in grams.
C) Next, consider a situation in which all of the S is consumed before all of the O2 reacts, or one in which you have excess S because all of the O2 has been used up.
For each of the given situations, indicate whether S or O2 is the limiting reactant.
- 3.00 mol Sulfur, 3.00 mol Oxygen
-3.00 mol Sulfur, 4.00 mol Oxygen
-3.00 mol Sulfur, 5.00 mol Oxygen
2S + 3O2 → 2SO3
General equation 2*32 = 64g 3*32 = 96g 2*80 = 160g
in this problem 4.13g (4.13*96)/64= 6.195g
Part A) 64g of sulfur require 96 g of O2 to produce 160g of SO3
4.13g of sulfur requires (4.13*96)/64 = 6.195g of O2
Moles = weight /molecular weight = 6.195/32 = 0.1936mole of O2
one mole = 6.023x1023 molecules
Part B) 64g of Sulfur can produce 160g of SO3
4.13g of sulfur produce (4.13*160)/64 = 10.325 g of SO3
0.1936 mole = 0.1936 * 6.023x1023 molecules = 1.166x1023 molecules
Part C) 2 mole S will completly react with 3 moles of O2
So 3 mole S require 4.5 moles of Oxygen
i) 3.00 mol Sulfur, 3.00 mol Oxygen ( limiting reagent is Oxygen )
ii) 3.00 mol Sulfur, 4.00 mol Oxygen ( limiting reagent is Oxygen )
iii) 3.00 mol Sulfur, 5.00 mol Oxygen ( limiting reagent is Sulfur )