How many litres of oxygen (at STP) are requir | vedclass

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(A) The balanced chemical equation for the combustion of benzene $(C_6H_6)$ is:$2C_6H_6(l) + 15O_2(g) \longrightarrow 12CO_2(g) + 6H_2O(l)$Calculate t

Vedclass · Accepted Answer

$84$

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(A) The balanced chemical equation for the combustion of benzene $(C_6H_6)$ is:$2C_6H_6(l) + 15O_2(g) \longrightarrow 12CO_2(g) + 6H_2O(l)$Calculate the molar mass of benzene $(C_6H_6)$:$M = (6 	imes 12) + (6 	imes 1) = 72 + 6 = 78 \ g/mol$From the stoichiometry of the reaction, $2 \ mol$ of $C_6H_6$ require $15 \ mol$ of $O_2$.Therefore, $78 \ g$ of $C_6H_6$ $(1 \ mol)$ requires $7.5 \ mol$ of $O_2$.At $STP$, $1 \ mol$ of gas occupies $22.4 \ L$.Volume of $O_2$ required for $78 \ g$ of $C_6H_6 = 7.5 	imes 22.4 \ L = 168 \ L$.For $39 \ g$ of $C_6H_6$ (which is $0.5 \ mol$):Volume of $O_2 = \frac{168 \ L}{2} = 84 \ L$.

How many litres of oxygen (at $STP$) are required for complete combustion of $39 \ g$ of liquid benzene?
(Atomic weights: $C=12, O=16, H=1$)

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How many litres of oxygen (at $STP$) are required for complete combustion of $39 \ g$ of liquid benzene?(Atomic weights: $C=12, O=16, H=1$)