(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 $L$)?

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On electrical decomposition of $150 \ mL$ dry and pure $O_2$,$10\%$ of $O_2$ gets changed to $O_3$. The volume of the gaseous mixture after the reaction and the volume of the remaining gas left after passing through turpentine oil will be,respectively:

What weight of $SO_2$ can be made by burning sulphur in $5.0 \ mol$ of oxygen?

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Element $A$ reacts with oxygen to form a compound $A_2O_3$. If $0.359 \ g$ of $A$ reacts to form $0.559 \ g$ of the compound,the atomic mass of $A$ is:

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