Element S reacts with O2 to form SO3 according to the reaction 2S+3O2-->2SO3. Ho

Question

Element S reacts with O2 to form SO3 according to the reaction 2S+3O2-->2SO3. How many O2 molecules are needed to eat with 5.88g of S? What is the theoretical yield of SO3 produced by the quantities described in the first question? next consider a situation in which all of the S is consumed before all 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.

3.0 mol sulfur 3.0 mol oxygen 40 mol oxygen 5.0mol oxygen 3.0 mol sulfur .D mol suilfur Limiting reactant is sulfur Limiting reactant is oxygen

Explanation / Answer

The reaction:

2S+3O2-->2SO3

Question 1.

How many O2 molecules are needed to eat with 5.88g of S?

mol of S --> mass/MW = 5.88/32 = 0.18375 mol of S

ratio is; 2 mol of S = 3 mol of O2

0.18375 mol --> 3/2*0.18375 = 0.275625 mol of O2

1 mol of O" = 6.022*10^23 molecules

0.275625 mol of O2 = --> (0.275625)(6.022*10^23) = 1.6598*10^23 molecules of O2 required

Questoin 2.

What is the theoretical yield of SO3 produced by the quantities described in the first question?

theoretical yield of SO3 -->

ratio is; 2 mol of S = 2 mol of SO3

0.18375 mol --> 2/2*0.18375 = 0.18375 mol of SO3

mass = mol*MW = (0.18375 )(80.066) = 14.71 g of SO3

Question 3.

Next consider a situation in which all of the S is consumed before all 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. (IMAGE)

2S+3O2-->2SO3

then..

case I

3 mol of S, 3 mol of O2...

O2 is limiiting the reaction

case II

3 mol of S, 4 mol of O2...

ratio --> 4/3 = 1.33 ;we need 1.5 thereofre

O2 is limiting the reaction

case III

3 mol of S, 5 mol of O2...

5/4 = 1.66; we need only 1.5 so

S is limiting the reaction

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