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28.0 \mathrm{L} of \mathrm{CO}_2 is produced on complete combustion of 16.8 \mathrm{~L} aseous mixture of ethene and methane at 25^{\circ} \mathrm{C} and 1 \mathrm{~atm}.Heat evolved during the combustion process is_________ \mathrm{kJ}.

Given : \Delta \mathrm{H}_{\mathrm{C}}\left(\mathrm{CH}_4\right)=-900 \mathrm{kJmol}^{-1}
\Delta \mathrm{H}_{\mathrm{c}}\left(\mathrm{C}_2 \mathrm{H}_4\right)=-1400 \mathrm{~kJ} \mathrm{~mol}^{-1}

Option: 1

847


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{\text{Moles of mixture }}=\frac{\mathrm{Pv}}{\mathrm{RT}}=\frac{1 \times 16.8}{0.0821 \times 298}=0.6866

\text{moles Moles of} \mathrm{CO}_2= \frac{1 \times 28}{0.0821 \times 298}=1.144 \mathrm{~mole}

\mathrm{CH}_4+2 \mathrm{O}_2 \longrightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}
x                                 x

\mathrm{C}_2 \mathrm{H}_4+3 \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}
(0.6866-x) \quad 2(0.686-x)

Total \mathrm{CO}_2 produced =1.144

x+2(0.6866-x)=1.144
x=1.3732-1.144
=0.2292

Moles of \mathrm{CH}_4=0.2292
\text { Moles of } \mathrm{C}_2 \mathrm{H}_4 =0.6866-0.2292
                                 =0.4574

Total Heat produced

=(900 \times 0.2292)+(0.4574 \times 1400)
=206.28+640.36=846.64

Posted by

himanshu.meshram

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