We're asked to calculate the actual yield of melamine (#"C"_3"N"_3"(NH"_2")"_3#), given some equations and its percent yield.

I'm going to assume both of these reactions go *to completion*.

Let's first convert the given mass of urea (#"CO(NH"_2")"_2#) to moles, using its molar mass, calculated to be #60.07# #"g/mol"#:

#124.5cancel("kg urea")((10^3cancel("g urea"))/(1cancel("kg urea")))((1color(white)(l)"mol urea")/(60.07cancel("g urea")))#

#= color(red)(2073# #color(red)("mol CO(NH"_2")"_2#

According to the first equation, there is a one-one molar ratio of urea to hydrogen cyanate (#"HNCO"#), so there is also #color(red)(2073# moles of #"HNCO"# in the reaction, and we'll use this number for our second reaction.

Now, let's use the coefficients of the *second* reaction equation to find the relative number of moles of melamine that form:

#color(red)(2073)color(white)(l)cancel(color(red)("mol HNCO"))((1color(white)(l)"mol melamine")/(6cancel("mol HNCO"))) = color(blue)(345.4# #color(blue)("mol melamine"#

Now, using the molar mass of melamine (calculated to be #126.15# #"g/mol"#) to find the *theoretical yield* of melamine:

#color(blue)(345.4)color(white)(l)cancel(color(blue)("mol melamine"))((126.15color(white)(l)"g melamine")/(1cancel("mol melamine")))#

#= color(purple)(43580# #color(purple)("g melamine"#

Finally to calculate the *actual yield* of melamine, we multiply this value by the percent yield (#73.0%#):

#(color(purple)(43580# #color(purple)("g melamine"))(73.0%) = color(orange)(31810# #color(orange)("g melamine")#

#= color(orange)(31.8# #color(orange)("kg melamine"#

The actual yield of melamine was therefore #31.8# #"kg"#.

This value in *moles* is simply the gram value divided by the molar mass of melamine:

#color(orange)(31810)cancel(color(orange)("g melamine"))((1color(white)(l)"mol melamine")/(126.15cancel("g melamine"))) = color(orange)(252# #color(orange)("mol melamine"#