How much liters of oxygen at $STP$ is required for complete combustion of $39g$ liquid benzene?
A.$11.2$
B.$22.4$
C.$42$
D.$84$
Answer
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Hint:We know that combustion is a process in which a substance reacts with oxygen from the air. When complete combustion of liquid benzene takes place, the following reaction will occur
$$2{C_6}{H_6}(l) + 15{O_2}(g) \to 12C{O_2}(g) + 6{H_2}O(g)$$
Formula Used
Mass of $1$ mole of benzene $ = $ $12 \times 6 + 1 \times 6$
The volume occupied by $1$ mole of oxygen $ = $ $22.4L$
Complete step-by-step answer:We know that combustion is a process in which a substance reacts with oxygen from the air. It is an exothermic reaction and is of two types that are incomplete and complete combustion. The products formed when complete combustion of a hydrocarbon takes place are carbon dioxide and water. So here complete combustion of benzene takes place. The following reaction will occur
$$2{C_6}{H_6}(l) + 15{O_2}(g) \to 12C{O_2}(g) + 6{H_2}O(g)$$
So here we can see that combustion of liquid benzene produces carbon dioxide and water and when this equation is balanced, we see that $2$ moles of benzene react with $15$ moles of oxygen at $STP$
Mass of $2$ moles of ${C_6}{H_6}$ $ = 2[12 \times 6 + 1 \times 6]g$
Mass of $2$ moles of ${C_6}{H_6} = 2 \times 78g$
Liters in $15$ moles of ${O_2} = 15 \times 22.4L$
Now we see that $2 \times 78g$ of ${C_6}{H_6}$ reacts with $15 \times 22.4L$of ${O_2}$ at $STP$ . To find how much ${O_2}$ is required for combustion of $39g$ ${C_6}{H_6}$ , we apply the unitary method
$ \Rightarrow 2 \times 78g$ of ${C_6}{H_6}$$ \to 15 \times 22.4L$ of ${O_2}$ at $STP$
$ \Rightarrow 39g$ of ${C_6}{H_6}$$ \to \dfrac{{39}}{{2 \times 78}} \times 15 \times 22.4L$ of ${O_2}$ at $STP$
Solving this, we get $84L$ of ${O_2}$ is required at $STP$
Thus the correct option is $D$.
Note:It is important to note that here oxygen is a gaseous compound, so we will multiply the stoichiometric coefficient of ${O_2}$ by $22.4L$, the volume occupied by $1$ mole of gaseous substance at standard temperature and pressure. Standard temperature and pressure mean ${0^0}C$ and $1$ atmospheric pressure.
$$2{C_6}{H_6}(l) + 15{O_2}(g) \to 12C{O_2}(g) + 6{H_2}O(g)$$
Formula Used
Mass of $1$ mole of benzene $ = $ $12 \times 6 + 1 \times 6$
The volume occupied by $1$ mole of oxygen $ = $ $22.4L$
Complete step-by-step answer:We know that combustion is a process in which a substance reacts with oxygen from the air. It is an exothermic reaction and is of two types that are incomplete and complete combustion. The products formed when complete combustion of a hydrocarbon takes place are carbon dioxide and water. So here complete combustion of benzene takes place. The following reaction will occur
$$2{C_6}{H_6}(l) + 15{O_2}(g) \to 12C{O_2}(g) + 6{H_2}O(g)$$
So here we can see that combustion of liquid benzene produces carbon dioxide and water and when this equation is balanced, we see that $2$ moles of benzene react with $15$ moles of oxygen at $STP$
Mass of $2$ moles of ${C_6}{H_6}$ $ = 2[12 \times 6 + 1 \times 6]g$
Mass of $2$ moles of ${C_6}{H_6} = 2 \times 78g$
Liters in $15$ moles of ${O_2} = 15 \times 22.4L$
Now we see that $2 \times 78g$ of ${C_6}{H_6}$ reacts with $15 \times 22.4L$of ${O_2}$ at $STP$ . To find how much ${O_2}$ is required for combustion of $39g$ ${C_6}{H_6}$ , we apply the unitary method
$ \Rightarrow 2 \times 78g$ of ${C_6}{H_6}$$ \to 15 \times 22.4L$ of ${O_2}$ at $STP$
$ \Rightarrow 39g$ of ${C_6}{H_6}$$ \to \dfrac{{39}}{{2 \times 78}} \times 15 \times 22.4L$ of ${O_2}$ at $STP$
Solving this, we get $84L$ of ${O_2}$ is required at $STP$
Thus the correct option is $D$.
Note:It is important to note that here oxygen is a gaseous compound, so we will multiply the stoichiometric coefficient of ${O_2}$ by $22.4L$, the volume occupied by $1$ mole of gaseous substance at standard temperature and pressure. Standard temperature and pressure mean ${0^0}C$ and $1$ atmospheric pressure.
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