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Harrizon [31]
3 years ago
7

C2H4(g) + H2(g) mc030-1.jpg C2H6(g) Which change would likely cause the greatest increase in the rate of the reaction?

Chemistry
2 answers:
amm18123 years ago
8 0
<h3><u>Answer;</u></h3>

increasing temperature and pressure

<h3><u>Explanation;</u></h3>
  • The reaction involves gases;

<em>C2H4(g) + H2(g) → C2H6(g)</em>

  • Therefore, <em><u>a change in pressure and temperature greatly affects the rate of reaction or the rate at which the products are formed.</u></em>
  • <em><u>Increasing the pressure in the reaction favors the rate of reaction</u></em> as the 2 volumes of the reactants forms 1 volume of the product and thus compression is evident, which will increase the rate of reaction.
  • <em><u>Increasing the temperature also increases the rate of reaction due to increases in the kinetic energy of molecules of the reactants and thus an increase in the number of collision by the particles of the reactants </u></em>and thus the speed at which the products are formed.
Vinil7 [7]3 years ago
5 0
The reaction is a hydrogenation reaction of an alkene, and its equation is:
C₂H₄(g) + H₂(g) → C₂H₆(g)
Therefore, this reaction can be sped up just as any other irreversible reaction may have its rate increased, by increasing temperature and pressure to increase the effective collisions of molecules.
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Solid aluminum (AI) and oxygen (O_2) gas react to form solid aluminum oxide (Al_2O_3). Suppose you have 7.0 mol of Al and 9.0 mo
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Answer:

There will be formed 3.5 moles of Al2O3 ( 356.9 grams)

Explanation:

Step 1: Data given

Numbers of Al = 7.0 mol

Numbers of mol O2 = O2

Molar mass of Al = 26.98 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

4Al(s) + 3O2(g) → 2Al2O3(s)  

Step 3: Calculate limiting reactant

For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3

Al is the limiting reactant, it will be consumed completely (7 moles).

O2 is in excess.  There will react 3/4 * 7 = 5.25 moles

There will remain 9-5.25 = 3.75 moles

Step 4: Calculate moles Al2O3

For 4 moles Al we'll have 2moles Al2O3

For 7.0 moles of Al we'll have 3.5 moles of Al2O3 produced

Step 5: Calculate mass of Al2O3

Mass Al2O3 = moles Al2O3 * molar mass Al2O3

Mass Al2O3 = 3.5 moles* 101.96 g/mol

Mass Al2O3 = 356.9 grams

There will be formed 3.5 moles of Al2O3 ( 356.9 grams)

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