# Consider the following reaction: 2N2(g) + 5O2(g) + 2H2O(g) --> 4HNO3(g). If a reaction mixture contains 28g of N2, 150g of O2, and 36g of H2O, what is the limiting reactant?

Nov 1, 2015

You have the equation:

${N}_{2} \left(g\right) + 5 {O}_{2} \left(g\right) + 2 {H}_{2} O \left(g\right) \rightarrow 4 H N {O}_{3}$

#### Explanation:

You have been given the masses of each reagent. These are molar quantiies when we divide them by the molecular mass of each reagent:

i.e. $\frac{28 \cdot g \cdot {N}_{2}}{28 \cdot g \cdot m o {l}^{-} 1}$; $\frac{150 \cdot g \cdot {O}_{2}}{32 \cdot g \cdot m o {l}^{-} 1}$; $\frac{36 \cdot g \cdot {H}_{2} O}{36 \cdot g \cdot m o {l}^{-} 1}$.

It doesn't too much arithmetic to realize that the in terms of molar quantities you have ${N}_{2} : {O}_{2} : {H}_{2} O : H N {O}_{3}$ $=$ $1 : < 5 : 2$. The reaction mixture is almost stoichiometric (what do I mean?), but there is a slight deficiency of oxygen gas. The limiting reagent is therefore ??.