Identify the limiting reactant in the reaction of carbon monoxide and oxygen to form co2, if 9.16 g of co and 9.01 g of o2 are combined. determine the amount (in grams) of excess reactant that remains after the reaction is complete.
formula of limiting reactant =
amount of excess reactant remaining = g

The formula of limiting reagent is 'CO' and amount of excess reagent remaining is 3.68 grams.

Explanation:

To calculate the number of moles, we use the equation

  ....(1)

Given mass of carbon monoxide = 9.16 g

Molar mass of carbon monoxide = 28 g/mol

Putting values in equation 1, we get:

Given mass of oxygen gas = 9.01 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

For the given chemical equation:

By Stoichiometry of the reaction:

2 mole of carbon monoxide reacts with 1 mole of oxygen gas

So, 0.33 moles of carbon monoxide will react with = of oxygen gas

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

So, carbon monoxide is considered as a limiting reagent because it limits the formation of products.

Now, calculating the mass of oxygen gas from equation 1, we get:

Moles of oxygen gas = 0.115 moles

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

Hence, the formula of limiting reagent is 'CO' and amount of excess reagent remaining is 3.68 grams.

the potential energy of the reactants is 200j.

from the energy diagram, the energy of the product formed is 350j; this means that, this reaction is an endothermic reaction, because it absorbs energy from its environment.