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pentagon [3]
1 year ago
10

Explain the problem surrounding the ammonia-making process in terms of chemical equilibrium.

Chemistry
1 answer:
ElenaW [278]1 year ago
4 0

The main problem is the release of heat and thereby reducing yield of Ammonia.

<h3>What is chemical equilibrium?</h3>

Chemical equilibrium is defined as the state of a reversible chemical reaction where there is no change in net amount of products and reactants involved in it.

What is an exothermic reaction?

Exothermic reaction is one of the chemical reactions that will produces heat during the formation of products.

At the chemical equilibrium, the chemical reaction will be as follows:

N_{2} + 3H_{2} → 2NH_{3} + energy

The above chemical equilibrium equation indicates that energy will be released due to the reaction of nitrogen and hydrogen. The energy release increases the surrounding temperature and thereby it will reduce the yield of the ammonia. Thus it can be indicated as the major problem.

To know more about exothermic reaction visit:

brainly.com/question/10373907

#SPJ4

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When 69.9 g heptane is burned it releases __ mol water.
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Answer:

1) When 69.9 g heptane is burned it releases 5.6 mol water.  

2) C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O.

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  • Firstly, we should balance the equation of heptane combustion.
  • The balanced equation is: <em>C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O.</em>

This means that every 1.0 mole of complete combustion of heptane will release 8 moles of H₂O.

  • We need to calculate the no. of moles in 69.9 g of heptane that is burned using the relation: <em>n = mass/molar mass.</em>

n of 69.9 g of heptane = mass/molar mass = (69.9 g)/(100.21 g/mol) = 0.697 mol ≅ 0.7 mol.

<em><u>Using cross multiplication:</u></em>

1.0 mol of heptane releases → 8 moles of water.

0.7 mol of heptane releases → ??? moles of water.

<em>∴ The no. of moles of water that will be released from burning (69.9 g) of water</em> = (0.7 mol)(8.0 mol)/(1.0 mol) = <em>5.6 mol.</em>

<em>∴ When 69.9 g heptane is burned it releases </em><em>5.6</em><em> mol water. </em>

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