<u>Answer:</u> 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.
<u>Explanation:</u>
To calculate the number of moles, we use the equation
....(1)
Given mass of ozone = 0.827 g
Molar mass of ozone = 48 g/mol
Putting values in above equation, we get:
![\text{Moles of ozone}=\frac{0.827g}{48g/mol}=0.0172mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20ozone%7D%3D%5Cfrac%7B0.827g%7D%7B48g%2Fmol%7D%3D0.0172mol)
Given mass of nitric oxide = 0.635 g
Molar mass of nitric oxide = 30.01 g/mol
Putting values in above equation, we get:
![\text{Moles of nitric oxide}=\frac{0.635g}{30.01g/mol}=0.0211mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20nitric%20oxide%7D%3D%5Cfrac%7B0.635g%7D%7B30.01g%2Fmol%7D%3D0.0211mol)
For the given chemical equation:
![O_3+NO\rightarrow O_2+NO_2](https://tex.z-dn.net/?f=O_3%2BNO%5Crightarrow%20O_2%2BNO_2)
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 =
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 =
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:
![0.0172mol=\frac{\text{Mass of nitrogen dioxide}}{46g/mol}\\\\\text{Mass of nitrogen dioxide}=0.7912g](https://tex.z-dn.net/?f=0.0172mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20nitrogen%20dioxide%7D%7D%7B46g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20nitrogen%20dioxide%7D%3D0.7912g)
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.