Answer to Question #141435 in General Chemistry for Sarah Dodd

Question

-How many O2 molecules are needed to react with 2.80 g of S?
-What is the theoretical yield of SO3 produced by 2.80 g of S?

Answer 1

Reaction of S with O₂ is :

 2S(g) + 3O₂(g) –––> 2SO₃(g)

  

According to above reaction

2 mole of S needs 3 moles of O₂

Molar mass of S = 32 g/mol

So, mass of 2 mole of S = 32 g/mol × 2 mole

                      = 64 g 

So, 

64 g of S needs 3 moles of O₂

1 g of S needs (3/64) mole of O₂

2.80 g of S needs (3/64)×2.80 mole of O2

               = 0.1313 mole O₂

Part 1            

In 1 mole of any substance number of molicule = 6.023×10²³ molicules

So, 0.1313 mole O₂

    = (0.1313×6.023×10^23) molicules

    = 7.908×10²² molicules

Hence,

7.908×10²² molicules O₂ molecules are needed to react with 2.80 g of S

Part 2

Theoriticly 

2.80 g of S needs 0.1313 mole O₂

Molar mass of O₂ = 32 g/mol

Mass of 0.1313 mole O₂ = 32×0.1313 g

                      = 4.202 g 

Hence,

Theoriticly 2.80 g of S needs 4.202 g O₂

So, theoriticl yield = 4.202 g