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The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 103 kJ heat by burning one of the fuels. Which fuel will emit the least amount of CO2? 1. Calculate the number of grams needed of each fuel: 2. Calculate the number of moles of each fuel: 3. Write down the balanced chemical equation for the combustion of the fuels: 4. Calculate the number of moles of CO2 produced by burning each fuel to produce 1.000 x 103 kJ. Which fuel will emit the least amount of CO2?

Respuesta :

Answer :

(1) The number of grams needed of each fuel [tex](C_2H_6)\text{ and }(C_4H_{10})[/tex] are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel [tex](C_2H_6)\text{ and }(C_4H_{10})[/tex] are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

[tex]C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O[/tex]

[tex]C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O[/tex]

(4) The number of moles of [tex]CO_2[/tex] produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of [tex]CO_2[/tex] is [tex]C_2H_6[/tex]

Explanation :

Part 1 :

First we have to calculate the number of grams needed of each fuel [tex](C_2H_6)\text{ and }(C_4H_{10})[/tex].

As, 52 kJ energy required amount of [tex]C_2H_6[/tex] = 1 g

So, 1000 kJ energy required amount of [tex]C_2H_6[/tex] = [tex]\frac{1000}{52}=19.23g[/tex]

and,

As, 49 kJ energy required amount of [tex]C_4H_{10}[/tex] = 1 g

So, 1000 kJ energy required amount of [tex]C_4H_{10}[/tex] = [tex]\frac{1000}{49}=20.41g[/tex]

Part 2 :

Now we have to calculate the number of moles of each fuel [tex](C_2H_6)\text{ and }(C_4H_{10})[/tex].

Molar mass of [tex]C_2H_6[/tex] = 30 g/mole

Molar mass of [tex]C_4H_{10}[/tex] = 58 g/mole

[tex]\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles[/tex]

and,

[tex]\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles[/tex]

Part 3 :

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of [tex]C_2H_6[/tex] is:

[tex]C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O[/tex]

and,

The balanced chemical reaction for combustion of [tex]C_4H_{10}[/tex] is:

[tex]C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O[/tex]

Part 4 :

Now we have to calculate the number of moles of [tex]CO_2[/tex] produced by burning each fuel to produce 1000 kJ.

[tex]C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O[/tex]

From this we conclude that,

As, 1 mole of [tex]C_2H_6[/tex] react to produce 2 moles of [tex]CO_2[/tex]

As, 0.641 mole of [tex]C_2H_6[/tex] react to produce [tex]0.641\times 2=1.28[/tex] moles of [tex]CO_2[/tex]

and,

[tex]C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O[/tex]

From this we conclude that,

As, 1 mole of [tex]C_4H_{10}[/tex] react to produce 4 moles of [tex]CO_2[/tex]

As, 0.352 mole of [tex]C_4H_{10}[/tex] react to produce [tex]0.352\times 4=1.41[/tex] moles of [tex]CO_2[/tex]

So, the fuel that emitting least amount of [tex]CO_2[/tex] is [tex]C_2H_6[/tex]

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