Question

Q.4. If 4.95 g of ethylene (C2H4) are combusted with 3.25 g of oxygen. a. What is the limiting reagent? b. How many grams of CO2 are formed?

Answer 1

Explanation:

For (a): The limiting reagent is oxygen gas.

For (b): The amount of carbon dioxide produced is 2.99 grams

Explanation:

For a:

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

.....(1)

For ethylene:

Given mass of ethylene = 4.95 g

Molar mass of ethylene = 28 g/mol

Putting values in equation 1, we get:

For oxygen gas:

Given mass of oxygen gas = 3.25 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

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

By Stoichiometry of the reaction:

3 moles of oxygen gas reacts with 1 mole of ethylene gas

So, 0.102 moles of oxygen gas will react with = of ethylene

As, given amount of ethylene gas 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 carbon dioxide

So, 0.102 moles of oxygen gas will produce = of carbon dioxide

For b:

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.068 moles

Putting values in equation 1, we get: