If 156 g of sodium nitrate react, and 112 g of sodium nitrite are recovered, what is the percentage yield?

1 Answer
Jun 21, 2017

Answer:

#%"yield" = 88.4%#

Explanation:

I'm going to assume the reaction is the following:

#2"NaNO"_3(s) rarr 2"NaNO"_2 (s) + "O"_2(g)#

with an supplement of heat energy to "drive" the reaction.

We're given that #156# #"g NaNO"_3# react, so let's convert this to moles using its molar mass, calculated to be #85.00"g"/"mol"#:

#156cancel("g NaNO"_3)((1color(white)(l)"mol NaNO"_3)/(85.00cancel("g NaNO"_3))) = 1.84# #"mol NaNO"_3#

Now, using the stoichiometrically equivalent values (the coefficients), let's determine the relative number of moles of #"NaNO"_2# that should form:

#1.84cancel("mol NaNO"_3)((2color(white)(l)"mol NaNO"_2)/(2cancel("mol NaNO"_3))) = 1.84# #"mol NaNO"_2#

Finally, let's use the molar mass of sodium nitrite (#69.00"g"/"mol"#) to calculate the theoretical mass yield:

#1.84cancel("mol NaNO"_2)((69.00color(white)(l)"g NaNO"_2)/(1cancel("mol NaNO"_2))) = color(red)(127# #color(red)("g NaNO"_2#

We're given that the actual yield of #"NaNO"_2# is #112# #"g"#, so we can use the percent yield equation

#%"yield" = "experimental yield"/"theoretical yield" xx 100%#

to find the percent yield:

#%"yield" = (112cancel("g NaNO"_2))/(color(red)(127)cancel(color(red)("g NaNO"_2))) xx 100% = color(blue)(88.4%#

The percent yield of sodium nitrite for this reaction is thus #color(blue)(88.4%#.