Question #6fe85

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

#color(blue)(2)"Na"_ ((s)) + "O"_ (2(g)) -> "Na"_ 2"O"_ (2(s))#

Notice that you need #1# mole of oxygen gas to react with #color(blue)(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 color(red)(cancel(color(black)("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 #"6.34 g"# of sodium peroxide. This is known as the reaction's theoretical yield.

However, you know that only #"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.