Question

100g nitrogen mixed with 92g hydrogen
a) identity the limiting reagent
b) calculate the weight of amonia form
c) which reactent remains and how much

Answer 1

n of N2 = 100/14=3.57 moles
n of H2= 92/2 =46 H2 is in access ,so N2 is
limiting reagent,and H2 remains (46-3*3.57=35.29)


3.57
2*3.57
N2 + 3H2 = 2NH3

m of NH3 is 2*3.57*17=121.38 g