The limitant reactant of a reaction is the reactant that we have the least number of mols considering its coefficient on the stoichiometry.
The reaction in this case is:
[tex]NH_3+5O_2\to4NO+6H_2O[/tex]The reactants are the one on the left side, so the answer can't be NO or H₂O.
The first thing to do is to calculate the number of moles of NH₃ and O₂, but for this we need their molar masses:
[tex]M_{NH_3}=1\cdot M_n+3\cdot M_H=(1\cdot14.0067+3\cdot1.00794)g/mol=17.03052g/mol[/tex][tex]M_{O_2}=2\cdot M_O=2\cdot15.9994g/mol=31.9988g/mol[/tex]Now, using them, let's calculate the number of moles of each:
[tex]\begin{gathered} M_{NH_{3}}=\frac{m_{NH_3}}{n_{NH_{3}}} \\ n_{NH_3}=\frac{m_{NH_3}}{M_{NH_{3}}}=\frac{2g}{17.03052g/mol}=0.11743\ldots.mol \end{gathered}[/tex][tex]\begin{gathered} M_{O_{2}}=\frac{m_{O_2}}{n_{O_{2}}} \\ n_{O_2}=\frac{m_{O_2}}{M_{O_{2}}}=\frac{4g}{31.9988g/mol}=0.12500\ldots mol \end{gathered}[/tex]Now, we can't compair directly, we need to consider their coefficients.
To do this, we divide the number of moles by the coefficient:
[tex]\begin{gathered} NH_3\colon\frac{n_{NH_3}}{1}=\frac{0.11743\ldots mol}{1}=0.11743\ldots mol \\ O_2\colon\frac{n_{O_{2}}}{5}=\frac{0.12500\ldots mol}{5}=0.025000\ldots mol \end{gathered}[/tex]Now, we can compair.
Since the value we have got to O₂ is less than NH₃, the limiting reactant is O₂
