Question

50.0 kg of N (9) and 10.0 kg of H (9) are mixed to produce NH3(g). Calculate the moles of
NH3(9) formed. Identify the limiting reagent in the production of NH, in this situation.
Hint : N2(g) + 3H2(g) -> 2NH3(g)
)​

Answer 1

Answer:

o.5 moles

Explanation:

N2 is 50 kg

so one moleculei of N = 25 kg

H2 is 10 kg

so.. one atom of H is 5 kg

mass of nh3 ..... 25+5*3

= 40 kg

molecular mass of nh3 is 18

no of moles = masss in g / molecular mass

.. 40*1000/18

2222 .22 moles

nearly equal to 2 moles