(C) The molar mass of benzene $(C_6H_6)$ is $(6 \times 12) + (6 \times 1) = 78 \ g/mol$.Number of moles of $C_6H_6 = \frac{39 \ g}{78 \ g/mol} = 0.5 \

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(C) The molar mass of benzene $(C_6H_6)$ is $(6 \times 12) + (6 \times 1) = 78 \ g/mol$.Number of moles of $C_6H_6 = \frac{39 \ g}{78 \ g/mol} = 0.5 \ mol$.From the balanced chemical equation, $1 \ mol$ of $C_6H_6$ requires $\frac{15}{2} \ mol$ of $O_2$.Therefore, $0.5 \ mol$ of $C_6H_6$ requires $\frac{15}{2} \times 0.5 = 3.75 \ mol$ of $O_2$.At $STP$, $1 \ mol$ of any gas occupies $22.4 \ L$.Volume of $O_2$ at $STP = 3.75 \ mol \times 22.4 \ L/mol = 84 \ L$.

Class 11 Chemistry Some Basic Concepts of Chemistry Chemical stoichiometry

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$C_6H_6$ (liq) + $\frac{15}{2}O_2$ $(g)$ $\rightarrow$ $6CO_2$ $(g)$ + $3H_2O$ (liq). Benzene burns in oxygen according to the above equation. What is the volume of oxygen (at $STP$) needed for complete combustion of $39 \ g$ of liquid benzene (in $L$)?

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A
$11.2$
B
$22.4$
C
$84$
D
$168$

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Some Basic Concepts of Chemistry

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Chemical stoichiometry

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$C_6H_6$ (liq) + $\frac{15}{2}O_2$ $(g)$ $\rightarrow$ $6CO_2$ $(g)$ + $3H_2O$ (liq). Benzene burns in oxygen according to the above equation. What is the volume of oxygen (at $STP$) needed for complete combustion of $39 \ g$ of liquid benzene (in $L$)?

Similar Questions

$A$ bottle of cold drink contains $200 \, mL$ of liquid with $0.1 \, M$ $CO_2$. If $CO_2$ behaves ideally,what will be the volume of $CO_2$ gas at $STP$ in the cold drink solution (in $, L$)?

How many grams of $HCl$ will be present in $150 \ mL$ of its $0.52 \ M$ solution?

The number of moles of $Br_2$ produced when two moles of potassium permanganate are treated with excess potassium bromide in aqueous acid medium is

For the reaction $4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$,what is the stoichiometric ratio of the products $Na_2CrO_4 : Fe_2O_3 : CO_2$?

When $1 \ dm^{3}$ of $CO_{2}$ gas is passed over hot coke, the volume of the gaseous mixture after the complete reaction at $STP$ becomes $1.4 \ dm^{3}$. The composition of the gaseous mixture at $STP$ is:

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