What is the weight of oxygen that is required | vedclass

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(A) The balanced chemical equation for the combustion of ethylene $(C_2H_4)$ is:$C_2H_4 + 3O_2 ightarrow 2CO_2 + 2H_2O$Molar mass of $C_2H_4 = (2

Vedclass · Accepted Answer

$9.6$

Vedclass · Answer

(A) The balanced chemical equation for the combustion of ethylene $(C_2H_4)$ is:$C_2H_4 + 3O_2 ightarrow 2CO_2 + 2H_2O$Molar mass of $C_2H_4 = (2 	imes 12) + (4 	imes 1) = 28 \ g/mol$ (or $28 \ kg/kmol$).Molar mass of $3O_2 = 3 	imes (2 	imes 16) = 96 \ g/mol$ (or $96 \ kg/kmol$).From the stoichiometry,$28 \ kg$ of $C_2H_4$ requires $96 \ kg$ of $O_2$ for complete combustion.Therefore,for $2.8 \ kg$ of $C_2H_4$,the weight of $O_2$ required is:$= \frac{96 \ kg}{28 \ kg} 	imes 2.8 \ kg = 9.6 \ kg$.

What is the weight of oxygen that is required for the complete combustion of $2.8 \ kg$ of ethylene? (in $kg$)

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