# In the reaction below, how do you identify the limiting agent when 0.47 moles of Sodium are reacted with 26g of Copper(II) Nitrate?

## $2 N a \left(s\right) + C u {\left(N {O}_{3}\right)}_{2} \to 2 N a N {O}_{3} + C u \left(s\right)$

Feb 10, 2017

#### Answer:

The limiting reactant is "Cu"("NO"_3)_2.

#### Explanation:

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.

${M}_{r} : \textcolor{w h i t e}{m m m m m m m m} 187.56$
$\textcolor{w h i t e}{m m m m m m} \text{2Na" + "Cu"("NO"_3)_2 → "2NaNO"_3 + "Cu}$
$\text{Mass/g:} \textcolor{w h i t e}{m m m m m m m l} 26$
$\text{Amt/mol:} \textcolor{w h i t e}{m l} 0.47 \textcolor{w h i t e}{m m} 0.139$
$\text{Divide by:} \textcolor{w h i t e}{m m} 2 \textcolor{w h i t e}{m m m m} 1$
$\text{Moles rxn:} \textcolor{w h i t e}{m} 0.225 \textcolor{w h i t e}{m l l} 0.139$

2. Identify the limiting reactant

An easy way to identify the limiting reactant is to calculate the "moles of reaction" each 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.

"Cu"("NO"_3)_2 is the limiting reactant because it gives the fewest moles of reaction.