# Question 3b5b9

Mar 3, 2016

$\text{800 g}$

#### Explanation:

The idea here is that you need to use the balanced chemical equation and the given masses of the two reactants to determine whether or not you're dealing with a limiting reagent.

The balanced chemical equation for this synthesis reaction looks like this

$\textcolor{red}{2} {\text{Na"_text((s]) + "Cl"_text(2(g]) -> color(blue)(2)"NaCl}}_{\textrm{\left(s\right]}}$

The two reactants will react in a $\textcolor{red}{2} : 1$ mole ratio, which means that the reaction will always consume twice as many moles of sodium metal than you have moles of chlorine gas taking part in the reaction.

Use the molar masses of sodium metal and of chlorine gas to determine how many moles of each you have

500color(red)(cancel(color(black)("g"))) * "1 mole Na"/(23.0color(red)(cancel(color(black)("g")))) = "21.74 moles Na"

500color(red)(cancel(color(black)("g"))) * "1 mole Cl"_2/(70.91color(red)(cancel(color(black)("g")))) = "7.051 moles Cl"_2

According to the aforementioned mole ratio, $7.051$ moles of chlorine gas would require

7.051color(red)(cancel(color(black)("moles Cl"_2))) * (color(red)(2)" moles Na")/(1color(red)(cancel(color(black)("mole Cl"_2)))) = "14.10 moles Na"

Since you have more sodium metal than you need, this reactant will be in excess. This is of course equivalent to saying that chlorine gas will act as a limiting reagent, i.e. it will determine how much sodium takes part in the reaction.

Now, notice that you have a $\textcolor{red}{2} : \textcolor{b l u e}{2}$ mole ratio between sodium metal and sodium chloride.

This tells you that the reaction will produce as many moles of sodium chloride as you have moles of sodium metal that are taking part in the reaction.

In this case, $14.10$ moles of sodium are actually taking part in the reaction, which means that you will end up with $14.10$ moles of sodium chloride.

To get the mass of sodium chloride in grams, use the compound's molar mass

14.10 color(red)(cancel(color(black)("moles NaCl"))) * "58.44 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = "824 g"#

Since you only provided one sig fig for the masses of the two reactants, the answer must be rounded to one sig fig

${m}_{N a C l} = \textcolor{g r e e n}{\text{800 g}}$