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.

Answers: 3

Show answers

Answers

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

Learn more: link

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.

Answer from: Quest

A) the enthralling of formation for pure element in its standard state is always positivehope this : )

Answer from: Quest

The problem here is that according to aufbau principle before 3d orbital 4s subshell comes.after filling 3p orbital with electrons you have 4 electrons left, you place two in 4s with opposite spin and the other two in 3d in any of its two orbitals of total five orbitals with the same spin.

Another question on Chemistry

Chemistry, 21.06.2019 21:30

If i make a solution by adding 83grams of sodium hydroxide to 750ml i’d water what is the molarity of sodium hydroxide

Answers: 1

Answer

Chemistry, 22.06.2019 01:00

Consider three unlabeled bottles, each contain small pieces of one of the following metals. - magnesium - sodium - silver the following reagents are used for identifying the metals. - pure water - a solution of 1.0 molar hcl - a solution of concentrated hno3 (a) which metal can be easily identified because it is much softer than the other two? describe a chemical test that distinguishes this metal from the other two, using only one of the reagents above. write a balanced chemical equation for the reaction that occurs. (b) one of the other two metals reacts readily with the hcl solution. identify the metal and write the balanced chemical equation for the reaction that occurs when this metal is added to the hcl solution. use the table of standard reduction potentials (attached) to account for the fact that this metal reacts with hcl while the other does not. (c) the one remaining metal reacts with the concentrated hno3 solution. write a balanced chemical equation for the reaction that occurs. (d) the solution obtained in (c) is diluted and a few drops of 1 m hcl is added. describe what would be observed. write a balanced chemical equation for the reaction that occurs.

Answers: 2

Answer

Chemistry, 22.06.2019 08:30

For each of the compounds below, show that the charges on the ions add up to zero. a. kbr b. cao c. li(2)o d. cacl(2) e. alcl(3)

Answers: 2

Answer

Chemistry, 22.06.2019 08:30

Which common material is an example of a polymer? (25 pts) a. steel b. plastic c. petroleum d. rubbing alcohol

Answers: 2

Answer

You know the right answer?

The haber-bosch process is a very important industrial process. in the haber-bosch process, hydrogen...