Question

The reaction of sulfur with oxygen is shown below. 2s(s) + 3o2(g) → 2so3(g) calculate the mass of sulfur trioxide (in g) produced when 100.0 g of each reactant is present.

Answer 1

Answer: The mass of sulfur trioxide produced will be 167 grams

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

• For sulfur:

Given mass of sulfur = 100.0 g

Molar mass of sulfur = 32.07 g/mol

Putting values in equation 1, we get:

`\text{Moles of sulfur}=(100.0g)/(32.07g/mol)=3.12mol`

• For oxygen gas:

Given mass of oxygen gas = 100.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

`\text{Moles of oxygen gas}=(100.0g)/(32g/mol)=3.125mol`

The given chemical equation follows:

`2S(s)+3O_2(g)\rightarrow 2SO_3(g)`

By Stoichiometry of the reaction:

3 moles of oxygen gas reacts with 2 moles of sulfur metal

So, 3.125 moles of oxygen gas will react with =
`(2)/(3)* 3.125=2.08mol` of sulfur metal

As, given amount of sulfur metal is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of oxygen gas produces 2 moles of sulfur trioxide.

So, 3.125 moles of oxygen gas will produce
`(2)/(3)* 3.125=2.08mol` of sulfur trioxide

Now, calculating the mass of sulfur trioxide from equation 1, we get:

Molar mass of sulfur trioxide = 80.1 g/mol

Moles of sulfur trioxide = 2.08 moles

Putting values in equation 1, we get:

`2.08mol=\frac{\text{Mass of sulfur trioxide}}{80.1g/mol}\\\\\text{Mass of sulfur trioxide}=(2.08mol* 80.1g/mol)=167g`

Hence, the mass of sulfur trioxide produced will be 167 grams

Answer 2

The mass of sulfur trioxide (in g) produced when 100 g of each reactant is present is 166.64 grams

calculation

find the moles of each reactant

moles = mass/molar mass

moles of SO2= 100 g/ 64 = 1.5625 moles

moles of O2= 100/32= 3.125 moles

oxygen is the limiting reagent therefore by use of mole ratio of O2:SO3 which is 3:2 the moles of SO3= 3.125 x2/3=2.083 moles

mass of SO3= molar mass x moles

= 2.083 moles x80 g/mol= 166.64 grams