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.