Question

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

Answer 1

Answer: 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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$   ....(1)

• For carbon monoxide:

Given mass of carbon monoxide = 9.16 g

Molar mass of carbon monoxide = 28 g/mol

Putting values in equation 1, we get:

$\text{Moles of carbon monoxide}=\frac{9.16g}{28g/mol}=0.33mol$

• For oxygen gas:

Given mass of oxygen gas = 9.01 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{9.01g}{32g/mol}=0.28mol$

For the given chemical equation:

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

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 = $\frac{1}{2}\times 0.33=0.165moles$ 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.

• Amount of excess reagent (oxygen gas) left = 0.28 - 0.165 = 0.115 moles

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:

$0.115mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=3.68g$

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