Question

1.51 g H2 is allowed to react with 10.3 g N2, producing 2.65 g NH3. What is the theoretical yield in grams for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units.

Answer 1

Molecular weights : N2=28, H2= 2 and NH3=17

Reaction between N2 and H2 is given by

The reaction is N2+3H2 --à 2NH3

The reaction states that 1 moles of N2 reacts with 3 moles of hydrogen to give 2 moles ofammonia

Since mole= mass/molecular weight

For 1 mole, mass= molecular weight

Moles : H2= 1.51/2= 0.755g/mole    N2= 10.3/28= 0.37 g/mole

Stoichiometric ratio of N2: H2 ( as per the reaction ) = 1:3

As per the given quantities, the ratio is 0.37: 0.755 = 1: 2.05

Hence H2 is the limiting reactant and limits the yield of NH3.

3 moles of hydrogen produces 2 moles of ammonia

0.755 moles of hydrogen gives 0.755*2/3=0.5033 moles of NH3= 0.5033*17=8.556gms of NH3