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

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:

$\text{Moles of sulfur dioxide}=\frac{40.86g}{32g/mol}=0.638mol$

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:

$\text{Moles of oxygen gas}=\frac{40.01g}{32g/mol}=1.250mol$

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

$SO_2+\frac{1}{2}O_2\rightarrow SO_3$    $\Delta H=-197kJ/mol$

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 = $\frac{0.5}{1}\times 0.638=0.319mol$ 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 = $\frac{197}{1}\times 0.638=125.6kJ$

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