# Question 6fe85

Mar 10, 2017

#### Answer:

${\text{6.34 g Na"_2"O}}_{2}$

#### Explanation:

You can get sodium peroxide by burning sodium metal in the presence of oxygen gas

$\textcolor{b l u e}{2} {\text{Na"_ ((s)) + "O"_ (2(g)) -> "Na"_ 2"O}}_{2 \left(s\right)}$

Notice that you need $1$ mole of oxygen gas to react with $\textcolor{b l u e}{2}$ moles of sodium in order to produce $1$ mole of sodium peroxide.

Start by converting the mass of sodium to moles by using the element's molar mass

3.74 color(red)(cancel(color(black)("g"))) * "1 mole Na"/(23.0color(red)(cancel(color(black)("g")))) = "0.1626 moles Na"

Now, you know that oxygen gas is in excess, which means that all the moles of sodium will take part in the reaction.

In theory, the reaction will produce

0.1626 color(red)(cancel(color(black)("moles Na"))) * ("1 mole Na"_2"O"_2)/(color(blue)(2)color(red)(cancel(color(black)("moles Na")))) = "0.0813 moles Na"_2"O"_2

To convert this to grams of sodium peroxide, use the compound's molar mass

$0.0813 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles Na"_2"O"_2))) * "77.98 g"/(1color(red)(cancel(color(black)("mole Na"_2"O"_2)))) = color(darkgreen)(ul(color(black)("6.34 g Na"_2"O}}_{2}}}}$

The answer is rounded to three sig figs.

So, you can say that the reaction should produce $\text{6.34 g}$ of sodium peroxide. This is known as the reaction's theoretical yield.

However, you know that only $\text{6.01 g}$ of sodium peroxide are recovered. This is the reaction's actual yield.

This means that the reaction does not have a 100% yield. In other words, not all the moles of the two reactants that could produce sodium peroxide actually end up producing sodium peroxide.

You can find the reaction's percent yield by going

"% yield" = "actual yield"/"theoretical yield" xx 100%

In this case, you will have

"% yield" = (6.01 color(red)(cancel(color(black)("g"))))/(6.34color(red)(cancel(color(black)("g")))) xx 100% = 94.8%#

This means that for every $100$ moles of sodium peroxide that could be produced by the reaction, only $94.8$ moles are actually produced.