# Sodium and chlorine gas react to produce sodium chloride. If "234 g" of sodium chloride is produced, how much sodium reacted?

May 20, 2017

$\text{92.1 g}$

#### Explanation:

You know that

$2 {\text{Na"_ ((s)) + "Cl"_ (2(g)) -> 2"NaCl}}_{\left(s\right)}$

and that the reaction produced $\text{234 g}$ of sodium chloride. Convert this to moles by using the compound's molar mass

234 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44 color(red)(cancel(color(black)("g")))) = "4.004 moles NaCl"

Now, you know that the sample of sodium metal reacted completely, which implies that it was completely consumed by the reaction.

In other words, you don't have to worry about the sample of chlorine gas because the fact that sample of sodium metal was completely consumed lets you know that the chlorine gas is not a limiting reagent.

This means that the reaction consumed

4.004 color(red)(cancel(color(black)("moles NaCl"))) * "2 moles Na"/(2color(red)(cancel(color(black)("moles NaCl")))) = "4.004 moles Na"

Convert this to grams by using the element's molar mass

$4.004 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Na"))) * "22.99 g"/(1color(red)(cancel(color(black)("mole Na")))) = color(darkgreen)(ul(color(black)("92.1 g}}}}$

The answer is rounded to three sig figs.

SIDE NOTE You can show that chlorine gas is not the limiting reagent by calculating the number of moles present in the sample

0.142 color(red)(cancel(color(black)("kg"))) * "1 mole Cl"_2/(35.453color(red)(cancel(color(black)("kg")))) = "4.005 moles Cl"_2

In order to produce $4.004$ moles of sodium chloride, you only need

4.004 color(red)(cancel(color(black)("moles NaCl"))) * "1 mole Cl"_2/(2color(red)(cancel(color(black)("moles NaCl")))) = "2.002 moles Cl"_2

Since you have more chlorine gas than you need, chlorine is in excess.

May 20, 2017

The mass of sodium that reacted is 92.1 g, or 0.0921 kg.

#### Explanation:

$\text{2Na(s) + Cl"_2("g")}$$\rightarrow$$\text{2NaCl(s)}$

In order to determine the mass of chlorine gas that reacted to produce 234 g NaCl, you work backwords, starting with NaCl and ending with Na.

The process will involve the following steps:

$\textcolor{red}{\text{given mass NaCl}}$$\rightarrow$color(red)("mol NaCl"$\rightarrow$$\textcolor{red}{\text{mol Na}}$$\rightarrow$color(red)("mass Na"

color(blue)("Determine the mol NaCl that reacted."

Multiply the color(red)("given mass of NaCl" by the inverse of its molar mass.

The molar mass of $\text{NaCl}$ is $\text{58.44 g/mol NaCl}$.
https://www.ncbi.nlm.nih.gov/pccompound?term=NaCl

234color(red)cancel(color(black)("g NaCl"))xx(1"mol NaCl")/(58.44color(red)cancel(color(black)("g Cl"_2)))="4.004 mol NaCl"

color(blue)("Determine the moles of Na that reacted."

Multiply mol $\text{NaCl}$ by multiplying $\text{mol NaCl}$ by the mol ratio between $\text{NaCl}$ and $\text{Na}$ from the balanced equation, so that $\text{mol NaCl}$ is cancelled.

4.004color(red)cancel(color(black)("mol NaCl"))xx(2"mol Na")/(2color(red)cancel(color(black)("mol NaCl")))="4.004 mol Na"

color(blue)("Determine the mass of Na that reacted."

Multiply mol $\text{Na}$ by its molar mass.

The molar mass of $\text{Na}$ is $\text{22.98976928 g/mol Na} .$ (Periodic table)

4.004color(red)cancel(color(black)("mol Na"))xx(22.98976928"g Na")/(1color(red)cancel(color(black)("mol Na")))="92.1 g Na"="0.0921 kg" (rounded to three significant numbers)