# How can I identify the limiting reactant when 43.25 g of cac2 reacts with 33.71 g of water to produce ca (oh) to and c2h2?

Dec 15, 2014

${\text{CaC}}_{2}$ is the limiting reagent.

#### Explanation:

The limiting reagent is $C a {C}_{2}$.

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

Notice that we have a $1 : 2$ mole ratio between $C a {C}_{2}$ and ${H}_{2} O$; that is, every mole of the former used requires 2 moles of the latter.

Since the quantities of both reactants are given, and knowing that their molar masses are $62.0 \frac{g}{m o l}$ (for $C a {C}_{2}$) and $18.0 \frac{g}{m o l}$ (for ${H}_{2} O$), we can determine the number of moles from

${n}_{{H}_{2} O} = {m}_{{H}_{2} O} / \left(m o l a r m a s {s}_{{H}_{2} O}\right) = \frac{33.71 g}{18 \frac{g}{m o l}} = 1.87$ moles

${n}_{C a {C}_{2}} = {m}_{C a {C}_{2}} / \left(m o l a r m a s {s}_{C a {C}_{2}}\right) = \frac{43.25 g}{64.0 \frac{g}{m o l}} = 0.68$ moles

Notice that the number of moles of $C a {C}_{2}$ determined would require $2 \cdot 0.68 = 1.36$ moles of ${H}_{2} O$, less than what we have in the reaction. Therefore, ${H}_{2} O$ is in excess, which means that $C a {C}_{2}$ is the limiting reagent.