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(A) The combustion reaction for ethylene is: $C_2H_4 + 3O_2 \longrightarrow 2CO_2 + 2H_2O \quad \Delta H = -1411 \, kJ \, mol^{-1}$.From the stoichiom

Vedclass · Accepted Answer

$26.46$

Vedclass · Answer

(A) The combustion reaction for ethylene is: $C_2H_4 + 3O_2 \longrightarrow 2CO_2 + 2H_2O \quad \Delta H = -1411 \, kJ \, mol^{-1}$.From the stoichiometry,$1 \, mol$ of $C_2H_4$ reacts with $3 \, mol$ of $O_2$ molecules,which is equal to $6 \, mol$ of $O$ atoms.Since $1411 \, kJ$ of heat corresponds to $6 \, mol$ of $O$ atoms,$6226 \, kJ$ of heat corresponds to $\frac{6 	imes 6226}{1411} \, mol$ of $O$ atoms.Calculating this gives: $\frac{37356}{1411} \approx 26.46 \, mol$ of $O$ atoms.Therefore,the number of oxygen atoms is $26.46 \, N_A$.

When a certain amount of ethylene is combusted,$6226 \, kJ$ of heat is released. If its enthalpy of combustion is $1411 \, kJ \, mol^{-1}$,the number of oxygen atoms consumed in the process at $STP$ will be ............... $N_A$.

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