<u>Answer:</u> The formula of limiting reagent is 'CO' and amount of excess reagent remaining is 3.68 grams.
<u>Explanation:</u>
To calculate the number of moles, we use the equation
....(1)
- <u>For carbon monoxide:</u>
Given mass of carbon monoxide = 9.16 g
Molar mass of carbon monoxide = 28 g/mol
Putting values in equation 1, we get:
- <u>For oxygen gas:</u>
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.
- 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:
Hence, the formula of limiting reagent is 'CO' and amount of excess reagent remaining is 3.68 grams.