A reaction of #53.1 "g"# of #"Na"# and #35.7 "g"# of #"Br"_2# yields #40 "g"# of #"NaBr"#. What is the percentage yield?

Question

A reaction of #53.1 "g"# of #"Na"# and #35.7 "g"# of #"Br"_2# yields #40 "g"# of #"NaBr"#.  What is the percentage yield?

Cameron Andrews · Accepted Answer

#16.8%#

We need to apply the following formula to determine a reaction's percentage yield:

#% "yield": "yield from reaction"/("theoretical yield")xx100#

Now we must work out the theoretical yield of #"NaBr"#

Writing a balanced equation for the reaction is the first step.

#2"Na" color(white)(a)+ color(white)(a) "Br"_2 color(white)(a)rarr color(white)(a)2"NaBr"#

As we can see, the molar ratio of #"Na " : " NaBr"# is #1:1#, so we shall use these two molecules to work out the theoretical yield.

We need to calculate how many moles of each reagent (in this case, sodium) we have now that we have the balanced equation.

You can use the following formula to calculate moles:

#"n" ("moles")=("m " ("mass"))/("M (atomic mass)")#

Regarding sodium:

#"n"=53.1/23=2.31#

Now that we have the moles of sodium, we can work out the moles of sodium bromide. Since the molar ratio is #1:1#, the rules of stoichiometry tell us that we also have #2.31# moles of #"NaBr"#.

We have the moles and the atomic mass of #"NaBr"#, but we don't have the mass. However, we can easily rearrange the formula for moles.

#"n"="m"/"M"#

#"m"="n"xx"M"=2.31xx103=237.93 " g"#

We can now calculate the percentage yield.

#% " yield"=40/237.93xx100=16.8%#

Mateo Aguilar · Answer

undefined The percentage yield of the reaction is approximately 58.1%.