# Sodium and chlorine react to form sodium chloride: 2Na (s) + Cl2 (g) ----> 2NaCl (s) For the reaction of 44.0g Na with 69.9 g Cl2, what is the limiting reactant and what is the theoretical yield of sodium chloride in grams?

Jun 15, 2017

The limiting reactant is $\text{Na}$. The theoretical yield is 112 g $\text{NaCl}$.

#### Explanation:

This is a limiting reactant problem.

We know that we will need a balanced equation with masses, molar masses, and moles of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and everything else below the formulas.

${M}_{r} : \textcolor{w h i t e}{m m m m l} 22.99 \textcolor{w h i t e}{m l l} 70.91 \textcolor{w h i t e}{m l l} 58.44$
$\textcolor{w h i t e}{m m m m m m l l} \text{2Na" + color(white)(m)"Cl"_2 → "2NaCl}$
$\text{Mass/g:} \textcolor{w h i t e}{m m l l} 44.0 \textcolor{w h i t e}{m m l} 69.9$
$\text{Amt/mol:} \textcolor{w h i t e}{m l} 1.914 \textcolor{w h i t e}{m l} 0.9858$
$\text{Divide by:} \textcolor{w h i t e}{m m} 2 \textcolor{w h i t e}{m m m m} 1$
$\text{Moles rxn:} \textcolor{w h i t e}{l l} 0.9569 \textcolor{w h i t e}{m l} 0.9858$

$\text{Moles of Na" = 44.0 color(red)(cancel(color(black)("g Na"))) × "1 mol Na"/(22.99 color(red)(cancel(color(black)("g Na")))) = "1.914 mol Na}$

${\text{Moles of Cl"_2 = 69.9 color(red)(cancel(color(black)("g Cl"_2))) × "1 mol Cl"_2/(70.91 color(red)(cancel(color(black)("g Cl"_2)))) = "0.9858 mol Cl}}_{2}$

2. Identify the limiting reactant

An easy way to identify the limiting reactant is to calculate the "moles of reaction" each reactant will give:

You divide the moles of each reactant by its corresponding coefficient in the balanced equation.

I did that for you in the table above.

$\text{Na}$ is the limiting reactant because it gives the fewest moles of reaction.

3. Calculate the theoretical moles of $\text{NaCl}$

$\text{Theoretical yield" = 1.914 color(red)(cancel(color(black)("mol Na"))) × ("2 mol NaCl")/(2 color(red)(cancel(color(black)("mol Na")))) = "1.914 mol NaCl}$

4. Calculate the theoretical yield of $\text{NaCl}$

$\text{Theoretical yield" = 1.914 color(red)(cancel(color(black)("mol NaCl"))) × ("58.44 g NaCl")/(1 color(red)(cancel(color(black)("mol NaCl")))) = "112 g NaCl}$