Chemistry, 09.09.2019 22:10 fiorentinologan4

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.

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Answer from: mandy9386

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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Answer from: froyg1234

6.46 grams (rounded to 3 sig. fig.)

Explanation:

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:

$\rm H_2$ and $\rm N_2$ .

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 $\rm H_2$ is the limiting reactant. How many moles of $\rm NH_3$ will be produced?

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

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

Look up the coefficient ratio between $\rm H_2$ and $\rm NH_3$ in the balanced equation:

$\displaystyle \frac{n(\mathrm{NH_3})}{n(\mathrm{H_2})} = \frac{2}{3}$ .

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

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

(2)

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

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

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

Look up the coefficient ratio between $\rm N_2$ and $\rm NH_3$ in the balanced equation:

$\displaystyle \frac{n(\mathrm{NH_3})}{n(\mathrm{N_2})} = 2$ .

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

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

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

The mass of that much $\rm NH_3$ is approximately 6.46 grams.

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