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?

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:

     .....(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:

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:

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

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 = 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 =

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

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