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Luda [366]
3 years ago
7

If one of the impurities in diesel fuel has the formula c2h6s, what products will be formed when it burns? enter a balanced chem

ical equation for the reaction. include physical states for all species.
Chemistry
2 answers:
inna [77]3 years ago
7 0
<span>If one of the impurities in diesel fuel has the formula c2h6s, then the products that will form would be carbon dioxide, water and sulfur dioxide. The balanced chemical reaction would be as follows:

</span>C2H6S(l)<span> + 9/2O2(g) = 2CO2(g) + 3H2O(v) + SO2(g)
</span>
Hope this answers the question.
Vilka [71]3 years ago
3 0

Answer:

C_2H_6S(l)+5O_2(g)\rightarrow 2CO_2(g)+3H_2O(g)+SO_3(g)

Explanation:

Hello,

In this case, complete combustions are chemical reactions transforming compounds into their respective highest-oxidized oxides, it means, that for the given compound, the carbon will have +4 as its oxidation state, forming carbon dioxide as well as sulfur which will have +6, forming sulfur trioxide. Do not forget water is also formed as an "oxide" of hydrogen. Thus, the aforementioned chemical reaction is:

C_2H_6S+O_2\rightarrow CO_2+H_2O+SO_3

Thus, since the impurity will be in same segregation state of the fuel, that is liquid, and oxygen and all the products are gaseous we have:

C_2H_6S(l)+O_2(g)\rightarrow CO_2(g)+H_2O(g)+SO_3(g)

Finally, by balancing it, we then obtain:

C_2H_6S(l)+5O_2(g)\rightarrow 2CO_2(g)+3H_2O(g)+SO_3(g)

Best regards.

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This question is incomplete, the complete question is;

The rate constant for the reaction 3A equals 4B is 6.00 × 10⁻³ L.mol⁻¹min⁻¹.

how long will it take the concentration of A to drop from 0.75 to 0.25M ?

from the unit of the rate constant we know it is a second reaction order

OPTIONS

a) 2.2×10^−3 min

b) 5.5×10^−3 min

c) 180 min

d) 440 min

e) 5.0×10^2 min

Answer:

it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

Explanation:

Given that;

Rate constant K =  6.00 × 10⁻³ L.mol⁻¹min⁻¹

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given that it is a second reaction order;

k = 1/t [ 1/A - 1/A₀]

kt = [ 1/A - 1/A₀]

t = [ 1/A - 1/A₀] / k

K is the rate constant(6.00 × 10⁻³)

A₀ is initial concentration( 0.75 )

A is final concentration(0.25)

t is time required = ?

so we substitute our values into the equation

t = [ (1/0.25) - (1/0.75)] / (6.00 × 10⁻³)

t = 2.6666 / (6.00 × 10⁻³)

t = 444.34 ≈ 440 min     {significant figures}

Therefore it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

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