Question

Consider the combination reaction between sulfur dioxide gas and oxygen gas to produce sulfur trioxide. Also consider that you have 40.86 g of sulfur dioxide gas and 40.01 g of oxygen gas. For this reaction, use LaTeX: \Delta Δ Hrxn = - 197 kJ/mol of product. If the reaction proceeds to completion, what is the maximum amount of heat (in kJ) that you could expect to generate?

Answer 1

Answer: Maximum amount of heat that you could expect to generate is 125.6 kJ

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 dioxide = 40.86 g

Molar mass of sulfur dioxide = 64 g/mol

Putting values in equation 1, we get:

`\text{Moles of sulfur dioxide}=(40.86g)/(32g/mol)=0.638mol`

For oxygen gas:

Given mass of oxygen gas = 40.01 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

`\text{Moles of oxygen gas}=(40.01g)/(32g/mol)=1.250mol`

The chemical equation for the reaction of sulfur and oxygen gas follows:

`SO_2+(1)/(2)O_2\rightarrow SO_3`
`\Delta H=-197kJ/mol`

By Stoichiometry of the reaction:

1 mole of sulfur dioxide gas reacts with 0.5 mole of oxygen gas

So, 0.638 moles of sulfur dioxide gas will react with =
`(0.5)/(1)* 0.638=0.319mol` of oxygen gas

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

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

By Stoichiometry of the reaction:

1 mole of sulfur dioxide gas gas produces= 197 J of heat

So, 0.638 moles of sulfur dioxide gas will produce =
`(197)/(1)* 0.638=125.6kJ`

Thus maximum amount of heat (in kJ) that you could expect to generate is 125.6