<u>Answer:</u> The theoretical yield of the reaction is 14.1 grams.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
......(1)
Given mass of nitrogen dioxide = 65.0 g
Molar mass of nitrogen dioxide = 46 g/mol
Putting values in equation 1, we get:
![\text{Moles of nitrogen dioxide}=\frac{65.0g}{46g/mol}=1.41mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20nitrogen%20dioxide%7D%3D%5Cfrac%7B65.0g%7D%7B46g%2Fmol%7D%3D1.41mol)
The chemical equation for the reaction of nitrogen dioxide and water follows:
![3NO_2+H_2O\rightarrow 2HNO_3+NO](https://tex.z-dn.net/?f=3NO_2%2BH_2O%5Crightarrow%202HNO_3%2BNO)
By Stoichiometry of the reaction:
3 moles of nitrogen dioxide produces 1 mole of NO
So, 1.41 moles of nitrogen dioxide will produce =
of NO
Now, calculating the mass of NO by using equation 1, we get:
Moles of NO = 0.47 moles
Molar mass of NO = 30 g/mol
Putting values in equation 1, we get:
![0.47mol=\frac{\text{Mass of NO}}{30g/mol}\\\\\text{Mass of NO}=(0.47mol\times 30g/mol)=14.1g](https://tex.z-dn.net/?f=0.47mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20NO%7D%7D%7B30g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20NO%7D%3D%280.47mol%5Ctimes%2030g%2Fmol%29%3D14.1g)
Hence, the theoretical yield of the reaction is 14.1 grams.