# If I combine 15.0 grams of calcium hydroxide with 75.0 mL of 0.500M HCl, what is the limiting reagent from the reaction?

Dec 14, 2014

The limiting reagent is HCl.

The reaction's balanced chemical equation is

$C a {\left(O H\right)}_{2 \left(a q\right)} + 2 H C l \left(a q\right) \to C a C {l}_{2} \left(a q\right) + 2 {H}_{2} {O}_{\left(l\right)}$

We can see that we have a $1 : 2$ mole ratio
between $C a {\left(O H\right)}_{2}$ and $H C l$; that is, every mole of $C a {\left(O H\right)}_{2}$ used in the reaction needs 2 moles of
$H C l$. We can determine the number of moles of $H C l$ by using its
molarity, $C = \frac{n}{V}$

${n}_{H C l} = C \cdot V = 0.500 M \cdot 75.0 \cdot {10}^{-} 3 L = 0.038$ moles

We can determine the number of $C a {\left(O H\right)}_{2}$ moles to be ($C a {\left(O H\right)}_{2}$'s molar mass is $74 \frac{g}{m o l e}$)

${n}_{C a {\left(O H\right)}_{2}} = \frac{m}{m o l a r m a s s} = \frac{15.0 g}{74 \frac{g}{m o l e}} = 0.20$ moles:

Taking into consideration the mole-to-mole ration, $0.20$ moles of $C a {\left(O H\right)}_{2}$ would have required $2 \cdot 0.20 = 0.40$ moles of $H C l$; we can clearly see that the number of $H C l$ moles available is almost 10 times smaller, $0.038$.

Therefore, the limiting reagent is $H C l$.

As a side note, notice that since $H C l$ is a strong acid, which means it dissociates completely into ${H}^{+}$ and $C {l}^{-}$, and since $C a C {l}_{2}$ is a soluble salt which dissociates into $C {a}^{2 +}$ and $C {l}^{-}$ anions, the reaction's complete ionic equation would be

$C a {\left(O H\right)}_{2 \left(a q\right)} + 2 {H}^{+} \left(a q\right) + 2 C {l}^{-} \left(a q\right) \to C {a}^{2 +} \left(a q\right) + 2 C {l}^{-} \left(a q\right) + 2 {H}_{2} {O}_{\left(l\right)}$

The net ionic equation would be (after removing the spectator ions)

$C a {\left(O H\right)}_{2 \left(a q\right)} + 2 {H}^{+} \left(a q\right) \to C {a}^{2 +} \left(a q\right) + 2 {H}_{2} {O}_{\left(l\right)}$