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3 years ago

C. Balance these fossil-fuel combustion reactions. (1 point)

Chemistry

1 answer:

storchak [24]3 years ago

5 0

The question is incomplete. The complete question is :

C. Balance these fossil-fuel combustion reactions. (1 point)

C8H18(g) + 12.5O2(g) → ____CO2(g) + 9H2O(g) + heat

CH4(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat

C3H8(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat

C6H6(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat

Solution :

C8H18(g) + 12.5O2(g) → __8__CO2(g) + 9H2O(g) + heat

When 1 part of octane reacts with 12.5 parts of oxygen, it gives 8 parts of carbon dioxide and 9 parts of water along with liberation of energy.

CH4(g) + __2__O2(g) → __1__CO2(g) + __2__H2O(g) + heat

When 1 part of methane reacts with 2 parts of oxygen, it gives 1 part of carbon dioxide and 2 parts of water along with liberation of energy.

C3H8(g) + __5__O2(g) → __3__CO2(g) + __4__H2O(g) + heat

When 1 part of propane reacts with 5 parts of oxygen, it gives 3 part of carbon dioxide and 4 parts of water along with liberation of energy.

C6H6(g) + __1/2__O2(g) → __6__CO2(g) + __3__H2O(g) + heat

When 1 part of propane reacts with 1/2 parts of oxygen, it gives 6 part of carbon dioxide and 3 parts of water along with liberation of energy.

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Answer:

a) $T_b=590.775k$

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Explanation:

From the question we are told that:

Moles of N2 $n=2.50$

Atmospheric pressure $P=100atm$

Temperature $t=20 \textdegree C$

$t = 20+273$

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Initial heat $Q=1.36 * 10^4 J$

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Where

$C_v=Heat\ Capacity \approx 20.76 J/mol/K$

$T_b-T_a=\frac{1.36 * 10^4 J}{2.5*20.76 }$

$T_b-293k=297.775$

$T_b=590.775k$

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$PV=nRT$

For v double

$T_c=2*590.775k$

$T_c=1181.55k$

Therefore

$PV=Wbc$

$Wbc=(2.20)(8.314)(1181_590.778)$

$Wbc=10805.7J$

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$W_t=0+1.08*10^4$

$W_t=1.08*10^4J$

Generally the equation for amount of heat added is mathematically given by

$Q=nC_p\triangle T$

$Q=2.20*2907*(1181.55-590.775)\\$

$Q=3.778*10^4J$

Generally the equation for change in internal energy of the gas is mathematically given by

$\triangle V=nC_v \triangle T$

$\triangle V=2.20*20.76*(1181.55-293)k$

$\triangle V=4.058*10^4J$

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