The Haber-Bosch process is a very important industrial process. In the Haber-Bosch process, hydrogen gas reacts with nitrogen gas to produce ammonia according to the equation 3H2(g)+N2(g)→2NH3(g) The ammonia produced in the Haber-Bosch process has a wide range of uses, from fertilizer to pharmaceuticals. However, the production of ammonia is difficult, resulting in lower yields than those predicted from the chemical equation. 1.15 g H2 is allowed to react with 9.93 g N2, producing 1.12 g NH3. Part A 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.

The Theoretical Yield is calculated from the amount of reactants that are available. The value of the theoretical yield is independent of the actual yield.

This reaction involves two reactants:

[tex]\rm H_2[/tex] and
[tex]\rm N_2[/tex].

This question stated the mass of both. Start by determining which of the two is the limiting reactant.

Relative atomic mass data from a modern periodic table:

H: 1.008;
N: 14.007.

(1)

Assume that [tex]\rm H_2[/tex] is the limiting reactant. How many moles of [tex]\rm NH_3[/tex] will be produced?

[tex]M(\mathrm{H_2}) = 2\times 1.008 = \rm 2.016\; g\cdot mol^{-1}[/tex].

[tex]\begin{aligned} n(\mathrm{H_2}) &= \frac{m}{M}\\ &= \rm \frac{1.15\; g}{2.016\; g\cdot mol^{-1}}\\&\approx \rm 0.570437\; mol \end{aligned}[/tex].

Look up the coefficient ratio between [tex]\rm H_2[/tex] and [tex]\rm NH_3[/tex] in the balanced equation:

[tex]\displaystyle \frac{n(\mathrm{NH_3})}{n(\mathrm{H_2})} = \frac{2}{3}[/tex].

[tex]\begin{aligned} \frac{n(\mathrm{NH_3})}{n(\mathrm{H_2})}\cdot {n(\mathrm{H_2})} &= \frac{2}{3}\times \rm 0.570437\; mol \\&\approx \rm 0.380291\; mol\end{aligned}[/tex].

In other words, if [tex]\rm H_2[/tex] is the limiting reactant, approximately [tex]\rm 0.380291\; mol[/tex] of [tex]\rm NH_3[/tex] will be produced.

(2)

Similarly, assume that [tex]\rm N_2[/tex] is the limiting reactant. How many moles of [tex]\rm NH_3[/tex] will be produced?

[tex]M(\mathrm{N_2}) = 2\times 14.007 = \rm 28.014\; g\cdot mol^{-1}[/tex].

[tex]\begin{aligned} n(\mathrm{N_2}) &= \frac{m}{M}\\ &= \rm \frac{9.93\; g}{28.014\; g\cdot mol^{-1}}\\&\approx \rm 0.354466\; mol \end{aligned}[/tex].

Look up the coefficient ratio between [tex]\rm N_2[/tex] and [tex]\rm NH_3[/tex] in the balanced equation:

[tex]\displaystyle \frac{n(\mathrm{NH_3})}{n(\mathrm{N_2})} = 2[/tex].

[tex]\begin{aligned} \frac{n(\mathrm{NH_3})}{n(\mathrm{N_2})}\cdot {n(\mathrm{N_2})} &= 2\times 0.354466\rm \; mol \\&\approx \rm 0.708931\; mol\end{aligned}[/tex].

In other words, if [tex]\rm N_2[/tex] is the limiting reactant, approximately [tex]\rm 0.708931\; mol[/tex] of [tex]\rm NH_3[/tex] will be produced.

In theory, only around [tex]\rm 0.570437\; mol[/tex] of [tex]\rm NH_3[/tex] will be produced. The reaction will run out of [tex]\rm H_2[/tex] before all [tex]\rm N_2[/tex] are consumed.

The mass of that much [tex]\rm NH_3[/tex] is approximately 6.46 grams.

pstnonsonjoku pstnonsonjoku · Answer 2 · 2021-10-05T16:15:08+02:00

The theoretical yield of the reaction is 6.51 g of ammonia.

The equation of the reaction is; 3H2(g)+N2(g)→2NH3(g)

Number of moles of H2 = 1.15 g/2 g/mol = 0.575 moles

Number of moles of N2 = 9.93 g/28 g/mol = 0.355 mole

We need to find the limiting reactant which is the reactant that yields the least amount of product.

Given that;

3 moles of H2 yields 2 moles of ammonia

0.575 moles of H2 yields 0.575 moles × 2 moles/3 moles

= 0.383 moles of ammonia

Also;

1 mole of N2 yields 2 moles of ammonia

0.355 mole of N2 yields 0.355 mole× 2 moles/ 1 mole of N2

= 0.71 moles of ammonia

The limiting reactant is hydrogen.

Theoretical yield of ammonia = 0.383 moles of ammonia × 17 g/mol = 6.51 g of ammonia

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