Question

In the reaction CaO(s)+2NH4Cl(aq)->2NH3(g)+H2O(g)+CaCl2(s)
a. What is the theoretical yield if 112g of CaO is reacted with 224 g NH4Cl?
b. What is the percent yield if 16.3g of NH3 is actually isolated?

Answer 1

a)

First we will calculate the number of moles here

moles of CaO = 112/56=2

moles of NH4Cl = 224/53.5 =4.2

as we can see that CaO is limiting reagent as the stochiometric coefficients of CaO and ammonia are in ratio 1:2

hence ammonia formed will be double of CaO = 2*2 =4 moles

converting it into mass= 4*17=68 grams

b)

yield % = $\frac{actual \ yield}{expected \ yield}\times100=\frac{16.3}{68}\times100=23.97\%$