Determine the limiting reactant when 30.0 g of propane, c3h8, is burned with 75.0 g of oxygen.

The limiting reagent is oxygen gas.

Explanation:

Limiting reagent is defined as the reactant that is present in less amount and it limits the formation of products.

Excess reagent is defined as the reactant which is present in large amount.

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

     .....(1)

Given mass of propane = 30.0 g

Molar mass of propane = 44.1 g/mol

Putting values in equation 1, we get:

Given mass of oxygen = 75.0 g

Molar mass of oxygen = 32 g/mol

Putting values in equation 1, we get:

The chemical equation for the combustion of propane follows:

By Stoichiometry of the reaction:

5 moles of oxygen gas reacts with 1 mole of propane.

So, 2.34 moles of oxygen gas will react with = of propane

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

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

Hence, the limiting reagent is oxygen gas.

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying

the answer is 0.8m.

by: nicholas