Question

Ammonia is produced by the reaction of nitrogen and hydrogen according to the equation N2(g) + 3H2(g) → 2NH3(g)

Calculate the mass of ammonia produced when 32.0 g of nitrogen react with 13.5 g of hydrogen. ___________NH3

Which is the excess reactant and how much of it will be left over when the reaction is complete? hydrogen or nitrogen

________ g

Answer 1

Sol.

Reaction :

N2(g) + 3H2(g) -----> 2NH3(g)

As Mass of N2 = 32  g

Molar Mass of N2 = 28 g/mol

So , Moles of N2 = 32 / 28 = 1.1428 mol

and , Mass of H2 = 13.5 g

Molar Mass of H2 = 2 g/mol

So , Moles of H2 = 13.5 / 2 = 6.75 mol  

As 1 mole of N2 combines with 3 moles of H2

So , 1.1428 moles of N2 combines with

= 1.1428 × 3 = 3.4284 moles of H2

But there are 6.75 moles of H2 . So , H2 is the excess reactant and N2 is the limiting reactant .

Now , 1 mole of limiting reactant ( N2 ) gives 2 moles of NH3

So , 1.1428 moles of N2 gives

= 1.1428 × 2 = 2.2856 moles of NH3  

As Molar Mass of NH3 = 17.031 g/mol

So , Mass of NH3 produced

= 2.2856 × 17.031 = 38.926 g

Also, Moles of excess reactant ( H2) remains

= Total moles of H2 - moles of H2 combines with N2

= 6.75 - 3.4284 = 3.3216 mol

As Molar Mass of H2 = 2 g/mol

So , Mass of H2 remains = 3.3216 × 2 =   6.643   g