# Question 319d5

Sep 9, 2017

#### Answer:

The reaction uses 0.20 mol of $\text{Mg}$.

#### Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

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

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

${M}_{r} : \textcolor{w h i t e}{m m m m l} 24.30 \textcolor{w h i t e}{m m m l} 70.91$
$\textcolor{w h i t e}{m m m m m m m} {\text{Mg"color(white)(m) + color(white)(mll)"Cl"_2color(white)(m) → "MgCl}}_{2}$
$\text{Mass/g} : \textcolor{w h i t e}{m m l l} 5.05 \textcolor{w h i t e}{m m m l} 14 \textcolor{w h i t e}{m m m m m} 20$
$\text{Amt/mol:} \textcolor{w h i t e}{m m} 0.2078 \textcolor{w h i t e}{m m l l} 0.197$
$\text{Divide by:} \textcolor{w h i t e}{m m m} 1 \textcolor{w h i t e}{m m m m m} 1$
$\text{Moles rxn:} \textcolor{w h i t e}{m l l} 0.2078 \textcolor{w h i t e}{m m l l} 0.197$

$\text{Moles of Mg" = 5.05color(red)(cancel(color(black)("g Mg"))) × ("1 mol Mg")/(24.30 color(red)(cancel(color(black)("g Mg")))) = "0.2078 mol Mg}$

${\text{Moles of Cl"_2 = 14 color(red)(cancel(color(black)("g Cl"_2))) × ("1 mol Cl"_2)/(70.91 color(red)(cancel(color(black)("g Cl"_2)))) = "0.197 mol Cl}}_{2}$

Step 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.

${\text{Cl}}_{2}$ is the limiting reactant because it gives the fewer moles of reaction.

(The person who did this reaction must have been a great chemist, because they got a 105 % yield of product. The maximum possible yield is 19 g of "MgCl"_2)#.

Step 3. Calculate the moles of $\text{Mg}$ used

$\text{Moles of Mg" = 0.197 color(red)(cancel(color(black)("mol Cl"_2))) × ("1 mol Mg")/(1 color(red)(cancel(color(black)("mol Cl"_2)))) = "0.20 mol Mg}$

Note: The answer can have only two significant figures, because that is all you gave for the mass of ${\text{Cl}}_{2}$.