# What is the limiting reactant when 7.81 of HCl reacts with 5.24 g of NaOH to produce NaCl and H_2O?

May 15, 2017

$N a O H$

#### Explanation:

For the reaction,

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

Clearly, there is $1 : 1$ stoichiometry.

$\text{Moles of HCl} = \frac{7.81 \cdot g}{36.5 \cdot g \cdot m o {l}^{-} 1} = 0.214 \cdot m o l .$

$\text{Moles of NaOH} = \frac{5.24 \cdot g}{40.0 \cdot g \cdot m o {l}^{-} 1} = 0.131 \cdot m o l .$

Clearly, $H C l$ is in excess. Note that $H C l$ is a room temperature gas (that has some great solubility in water). Normally, an aqueous solution of a known concentration would be proposed for the acid, and you would access the molar quantity of $H C l$ by the relation:

$\text{Concentration"="Moles of HCl"/"Vollume of solution}$