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slavikrds [6]
3 years ago
12

Consider the combustion of octane (C8H18)

Chemistry
1 answer:
sesenic [268]3 years ago
3 0
The balanced equation for the above reaction is as follows;
2C₈H₁₈ + 25O₂ --->  16CO₂  + 18H₂O
stoichiometry of octane to CO₂ is 2:16
number of C₈H₁₈ moles reacted - 191.6 g / 114 g/mol = 1.68 mol
when 2 mol of octane reacts it forms 16 mol of CO₂
therefore when 1.68 mol of octane reacts - it forms 16/2 x 1.68 = 13.45 mol of CO₂
number of CO₂ moles formed - 13.45 mol
therefore mass of CO₂ formed - 13.45 mol x 44 g/mol = 591.8 g
 mass of CO₂ formed is 591.8 g
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Hello!
blondinia [14]

Answer:

[CO]=[Cl_2]=0.01436M

[COCl_2]=0.00064M

Explanation:

Hello there!

In this case, according to the given chemical reaction at equilibrium, we can set up the equilibrium expression as follows:

K=\frac{[CO][Cl_2]}{[COCl_2]}

Which can be written in terms of x, according to the ICE table:

0.32=\frac{x^2}{0.015M-x}

Thus, we solve for x to obtain that it has a value of 0.01436 M and therefore, the concentrations at equilibrium turn out to be:

[CO]=[Cl_2]=0.01436M

[COCl_2]=0.015M-0.01436M=0.00064M

Regards!

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