Question

How many milliliters of oxygen are required to react completely with 175 mL of ${{C}_{4}}{{H}_{10}}$ if the volumes of both gases are measured at the same temperature and pressure? The reaction is $2{{C}_{4}}{{H}_{10(g)}}+13{{O}_{2(g)}}\xrightarrow{{}}8C{{O}_{2(g)}}+10{{H}_{2}}{{O}_{(g)}}$

Answer 1

Hint: As it is said in the question that volumes of both gases are measured at the same temperature and pressure, which means we can use the ideal gas law equation to calculate volume of oxygen required to react with 175mL of butane.
Formula used:
We will use ideal gas equation i.e., $PV=nRT$

Complete answer:
The empirical relationship between volume, the temperature, the pressure, and the amount of a gas can be combined into the ideal gas law which can be written as follows:-
$PV=nRT$
Where,
P = pressure of the gas
V =volume of gas it occupies
n = number of moles
R = universal gas constant, equal to $0.0821\dfrac{atm\cdot L}{mol\cdot K}$
T = absolute temperature of the gas
 Now, let us see the reaction:-
$2{{C}_{4}}{{H}_{10(g)}}+13{{O}_{2(g)}}\xrightarrow{{}}8C{{O}_{2(g)}}+10{{H}_{2}}{{O}_{(g)}}$
We can very well conclude that butane and oxygen react in 2:13 mole ratio i.e., $\dfrac{{{n}_{bu\tan e}}}{{{n}_{oxygen}}}=\dfrac{2}{13}$
On rearranging it, we get: ${{n}_{butan e}}=\dfrac{2}{13}{{n}_{oxygen}}$
Volume of butane gas = 175mL
Let us consider constant temperature and pressure as T and P respectively. Then,
$\begin{align}
  & P{{V}_{Oxygen}}={{n}_{Oxygen}}RT\text{ -----eq(1)} \\
 & P{{V}_{Bu\tan e}}={{n}_{Bu\tan e}}RT\text{ -----eq(2)} \\
\end{align}$
On dividing equation (1) by (2), we get:-
\[=\dfrac{P{{V}_{Oxygen}}}{P{{V}_{Bu\tan e}}}=\dfrac{{{n}_{Oxygen}}RT}{{{n}_{Bu\tan e}}RT}\]
\[=\dfrac{{{V}_{Oxygen}}}{{{V}_{Bu\tan e}}}=\dfrac{{{n}_{Oxygen}}}{{{n}_{Bu\tan e}}}\]
On substituting the values, we get:-
\[=\dfrac{{{V}_{Oxygen}}}{175mL}=\dfrac{{{n}_{Oxygen}}}{\dfrac{2}{13}{{n}_{Oxygen}}}\]
\[=\dfrac{{{V}_{Oxygen}}}{175mL}=\dfrac{13}{2}\]
\[={{V}_{Oxygen}}=\dfrac{13}{2}\times 175mL\]
\[={{V}_{Oxygen}}=1137.5mL\]

Therefore, 1137.5mL of oxygen are required to react completely with 175mL of ${{C}_{4}}{{H}_{10}}$

Note:
While solving the question, write down the values and conditions provided. Always check the units and in order to avoid mistakes, try to solve the entire question along with units (preferably convert all the values to the same units before moving to further calculations).