Respuesta :
Answer: Nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.
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 ozone:
Given mass of ozone = 0.827 g
Molar mass of ozone = 48 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of ozone}=\frac{0.827g}{48g/mol}=0.0172mol[/tex]
- For nitric oxide:
Given mass of nitric oxide = 0.635 g
Molar mass of nitric oxide = 30.01 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of nitric oxide}=\frac{0.635g}{30.01g/mol}=0.0211mol[/tex]
For the given chemical equation:
[tex]O_3+NO\rightarrow O_2+NO_2[/tex]
By Stoichiometry of the reaction:
1 mole of ozone reacts with 1 mole of nitric oxide.
So, 0.0172 moles of ozone will react with = [tex]\frac{1}{1}\times 0.0172=0.0172moles[/tex] of nitric oxide
As, given amount of nitric oxide is more than the required amount. So, it is considered as an excess reagent.
Thus, ozone is considered as a limiting reagent because it limits the formation of product.
- Amount of excess reagent (nitric oxide) left = 0.0211 - 0.0172 = 0.0039 moles
By Stoichiometry of the reaction:
1 mole of ozone produces 1 mole of nitrogen dioxide.
So, 0.0172 moles of ozone will react with = [tex]\frac{1}{1}\times 0.0172=0.0172moles[/tex] of nitrogen dioxide
Now, calculating the mass of nitrogen dioxide from equation 1, we get:
Molar mass of nitrogen dioxide = 46 g/mol
Moles of nitrogen dioxide = 0.0172 moles
Putting values in equation 1, we get:
[tex]0.0172mol=\frac{\text{Mass of nitrogen dioxide}}{46g/mol}\\\\\text{Mass of nitrogen dioxide}=0.7912g[/tex]
Hence, nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.
