Question

Calculate the number of oxygen atoms required to combine with 7 g of $ {{\text{N}}_{\text{2}}} $ to form $ {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{3}}} $ when 80% of $ {{\text{N}}_{\text{2}}} $ is converted to $ {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{3}}} $.
 $ \text{2N}{{\text{O}}_{\text{2}}}\text{+3}{{\text{O}}_{\text{2}}}\to \text{2}{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{3}}} $

Answer 1

Hint: This question is based on stoichiometry of the reaction which refers to the number of molecules of the reactants and products involved in the balanced rate equation. We shall calculate the moles of nitrogen converted to nitrogen oxide and using the reaction equation, calculate the moles and thus particles of oxygen required.

Complete step by step solution:
From the given chemical equation we can see that,
 $ \text{2}{{\text{N}}_{\text{2}}}\text{+3}{{\text{O}}_{\text{2}}}\to \text{2}{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{3}}} $
2 moles of nitrogen reacts with 3 moles of oxygen to form two moles of dinitrogen trioxide.
As per the definition of mole, one mole of nitrogen is equal to the gram molecular weight of nitrogen molecules. Therefore, 28 g of nitrogen = 1 mole of nitrogen gas.
Therefore, 7 gram of nitrogen = $ \dfrac{1}{28}\times 7=0.25 $ mole.
However only 80% of $ {{\text{N}}_{\text{2}}} $ is converted to $ {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{3}}} $ = $ \dfrac{80}{100}\times 0.25=0.2 $ mole.
Now, $ 0.2 $ mole of nitrogen would require $ \dfrac{3}{2}\times 0.2=0.3 $ moles of oxygen gas.
Now, we need to find out the number of oxygen atoms required for the reaction and one mole of electrons is equal to $ 6.023\times {{10}^{23}} $ atoms. As there are $ 0.3 $ moles of oxygen gas this is equal to $ 6.023\times {{10}^{23}}\times 0.3=1.8069\times {{10}^{23}} $ atoms of oxygen.
Therefore, the number of oxygen atoms required to combine with 7 g of $ {{\text{N}}_{\text{2}}} $ to form $ {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{3}}} $ when 80% of $ {{\text{N}}_{\text{2}}} $ is converted to $ {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{3}}} $ is $ 1.8069\times {{10}^{23}} $ atoms of oxygen.

Note:
The reaction between nitrogen and oxygen to form different nitrogen oxides takes place in the atmosphere and at very high temperatures such as when the lightning strikes. Initially nitric oxide is formed which then subsequently forms the other oxides too on reaction with oxygen.