Question

50 g nitrogen gas and 8 g hydrogen gas are combined to produce NH₃ gas. Which is the limiting reagent of reaction? Calculate the amount of NH₃ gas produced. 

Answer 1

Balanced chemical equation is

Number of moles of nitrogen = \(50\ g\times \frac{1\ mol}{28\ g} = 1.786\ mol\)

Number of moles of hydrogen \( = 8\ g\ \times \ \frac{1\ mol}{2\ g} =4\ mol\) 

According to equation, 

Number of moles of H₂ reacts with 1.786 mol N₂ 

= 3/1 × 1.786 = 5.358 moles 

But moles present in H₂ = 4 

So complete H₂(g) will take part in reaction. So, H₂ iş limiting reagent. 

Amount of NH₃ (g) formed by 3 mol of H₂ = 2 mol 

Ammonia formed by 4 mol of H₂ = \(\frac{2}{3}\times 4\ mol\) 

= 2.66 mol 

2.66 mol = 2.66 × 17 =  45.22 g