Question

Elemental S reacts with O2 to form SO3 according to the reaction:

2S+ 3O2--> 2SO3

a.) How many molecules are needed to react with 5.74g of S? Express your answer numerically in units of molecules.

b.) What is the theoretical yield of SO3 produced by 5.74 g of S? Express your answer numerically in grams.

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.

c.) For each of the given situations, indicate whether S or O2 is the limiting reactant. Drag each item to the appropriate bin.

Bins: Limiting reactant is sulfur, and Limiting reactant is oxygen

Choices:

   - 3.0 mol sulfur, 4.0 mol of oxygen

   - 3.0 mol sulfur, 3.0 mol oxygen

   -3.0 mol sulfur, 5.0 mol oxygen

Answer 1

2S+ 3O2--> 2SO3

mass of S used = 5.74 g
molar mass of S= 32.07 g/mol
mols of S reacted = 5.74 g/32.07 g/mol= 0.17898 mols
a)
mol ratio S:O2 = 2:3
mols of O2 needed= [3/2]*0.17898 mols = 0.26847 mols
1 mol O2 = 6.022 x10^23 molecules of O2

0.26847 mols O2 = 0.26847 *6.022 x10^23 molecules of O2
= 1.617 x10^23 molecules of O2
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b)

mol ratio S: SO3 = 2:2 = 1:1

So mols of So3 produced = mols of S present = 0.17898 mols

molar mass of SO3=80.06 g/mol

mass of SO3 produced = 80.06 g/mol*0.17898 mols = 14.33 g

Theoretical yield of SO3 = 14.33 g
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2S+ 3O2--> 2SO3

mol ratio S : O2 = 2 :3

The reactant which consumed completely is the limiting reactant
3.0 mol sulfur, 4.0 mol of oxygen

3.0 mol sulfur need [3/2]* 3 mol O2 = 4.5 mols

but mols of O2 is fewer than this Only 4 mol O2 is available

So limiting reactant = O2
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3.0 mol sulfur, 3.0 mol oxygen

3.0 mol sulfur need [3/2]* 3 mol O2 = 4.5 mols but mols of O2 is fewer than this
Only 3 mol O2 is available

So limiting reactant = O2
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-3.0 mol sulfur, 5.0 mol oxygen

3.0 mol sulfur need [3/2]* 3 mol O2 = 4.5 mols
Here mols of O2 present = 5 mols, so O2 will be excess reactant

So limiting reactant =S
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Thanks ! Hope it helps !