Question

Which is the limiting reactant in the following chemical reaction: NH3 + 5 O2 —> 4 NO + 6 H2O, if 4 grams of O2 is used and 2 grams of NH3 is used ? Show work

Answer 1

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:

`NH_3+5O_2\to4NO+6H_2O`

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:

`M_(NH_3)=1\cdot M_n+3\cdot M_H=(1\cdot14.0067+3\cdot1.00794)g/mol=17.03052g/mol`
`M_(O_2)=2\cdot M_O=2\cdot15.9994g/mol=31.9988g/mol`

Now, using them, let's calculate the number of moles of each:

`\begin{gathered} M_{NH_(3)}=\frac{m_(NH_3)}{n_{NH_(3)}} \\ n_(NH_3)=\frac{m_(NH_3)}{M_{NH_(3)}}=(2g)/(17.03052g/mol)=0.11743\ldots.mol \end{gathered}`
`\begin{gathered} M_{O_(2)}=\frac{m_(O_2)}{n_{O_(2)}} \\ n_(O_2)=\frac{m_(O_2)}{M_{O_(2)}}=(4g)/(31.9988g/mol)=0.12500\ldots mol \end{gathered}`

Now, we can't compair directly, we need to consider their coefficients.

To do this, we divide the number of moles by the coefficient:

`\begin{gathered} NH_3\colon(n_(NH_3))/(1)=(0.11743\ldots mol)/(1)=0.11743\ldots mol \\ O_2\colon\frac{n_{O_(2)}}{5}=(0.12500\ldots mol)/(5)=0.025000\ldots mol \end{gathered}`

Now, we can compair.

Since the value we have got to O₂ is less than NH₃, the limiting reactant is O₂