Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia. N 2 ( g ) + 3 H 2 ( g ) ⟶ 2 NH 3 ( g ) Assume 0.180 mol N2 and 0.575 mol H2 are present initially. After complete reaction, how many moles of ammonia are produced

The answer to your question is 0.36 moles of NH₃

Explanation:

Data

moles of N₂ = 0.180

moles of H₂ = 0.575

moles of NH₃ = ?

- Balanced chemical reaction

             N₂ (g)  +  3H₂ (g)  ⇒  2NH₃ (g)

Process

1.- Calculate the limiting reactant

Theoretical proportion    N₂/H₂ = 1/3 = 0.333

Experimental proportion N₂/H₂ = 0.18 / 0.575 = 0.31

Conclusion

As the experimental proportion was lower than the theoretical proportion, we conclude that the N₂ is the limiting reactant.

2.- Calculate the moles of Ammonia

                    1 mol of N₂   2 moles of NH₃

                    0.180 moles  x

                    x = (0.180 x 2) / 1

                    x = 0.36 moles of NH₃

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

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