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

it is a good conductor

explanation:

Answer from: Quest

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.

Another question on Chemistry

Chemistry, 22.06.2019 08:00

An electron moved from shell n = 2 to shell n = 1. what most likely happened during the transition? a fraction of a photon was added. a photon of energy was absorbed. a fraction of a photon was removed. a photon of energy was released.

Answers: 1

Answer

Chemistry, 22.06.2019 16:30

4. a 20-kg child is tossed up into the air by her parent. the child is 2 meters off the ground traveling 5 m/s. circle one: ke / gpe / both show your work for finding the values of each type of energy the object has:

Answers: 1

Answer

Chemistry, 23.06.2019 02:00

What is the source of continuous heat and energy that we receive from the sun

Answers: 2

Answer

Chemistry, 23.06.2019 06:40

A250 g sample of water with an initial temperatureof 98.8 closes 6500 joules of heat. what is the finaltemperature of the water?

Answers: 1

Answer

You know the right answer?

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