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"#
