Respuesta :
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
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ....(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:
[tex]\text{Moles of carbon monoxide}=\frac{9.16g}{28g/mol}=0.33mol[/tex]
- 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:
[tex]\text{Moles of oxygen gas}=\frac{9.01g}{32g/mol}=0.28mol[/tex]
For the given chemical equation:
[tex]2CO(g)+O_2(g)\rightarrow 2CO_2(g)[/tex]
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 = [tex]\frac{1}{2}\times 0.33=0.165moles[/tex] 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:
[tex]0.115mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=3.68g[/tex]
Hence, the formula of limiting reagent is 'CO' and amount of excess reagent remaining is 3.68 grams.
