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Serhud [2]
3 years ago
6

An internal combustion engine relies primarily on the efficient production of energy by the combination of oxygen (02) and gasol

ine (assume this is primarily octane, C8H18). Write the chemical equation for this reaction and derive the exact ratio of oxygen to octane for 100% efficiency.
Chemistry
1 answer:
jeka57 [31]3 years ago
7 0
Answer is: <span>the exact ratio of oxygen to octane for is 12.5 : 1.
</span>Balanced chemical reaction: C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O or multiply by 2:
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O.
There same number of atoms on both side of balanced chemical reaction: eight carbon atoms, eighteen hydrogen atoms and twenty five oxygen atoms.

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Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =
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<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{c}

For a general chemical reaction:

aA+bB\rightleftharpoons cC+dD

The expression for K_{eq} is written as:

K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)

The expression of K_c for above equation is:

K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}

We are given:

[H_2S]_{eq}=0.671M

[O_2]_{eq}=0.587M

K_c=1.35

Putting values in above expression, we get:

1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}

[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M

Hence, the equilibrium concentration of water is 0.597 M

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When stress is added to plates they can diverge, converge, or slip past one another True or False
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