What volume of oxygen at 423.0 K and a pressure of 127.4 kPa is produced by the decomposition of 129.7g of BaO₂ to BaO and O₂?

The balanced equation is as follows –

\[{\rm{2Ba}}{{\rm{O}}_{\rm{2}}} \to {\rm{2BaO + }}{{\rm{O}}_{\rm{2}}}\]

Molecular weight of \[{\rm{Ba}}{{\rm{O}}_{\rm{2}}}{\rm{=AtomicweightofBa+2(Atomic weightofO)}}\]

\begin{aligned}{\rm{=137}}{\rm{.33+2(16)}}\\{\rm{=137}}{\rm{.33+32}}\\{\rm{=169}}{\rm{.33gm}}\end{aligned}

Molecular weight of \[{\rm{BaO = Atomic weight of Ba + Atomic weight of O}}\]

\begin{aligned}{l}{\rm{=137}}{\rm{.33+16}}\\{\rm{=153}}{\rm{.33 gm}}\end{aligned}

Molecular weight of \[{{\rm{O}}_2}{\rm{ = 2(Atomic weight of O)}}\]

\begin{aligned}{\rm{=2(16)}}\\{\rm{ = 32 gm}}\end{aligned}

So, the total moles according to the reaction is –

\begin{aligned}{\rm{2(169}}{\rm{.33)gm}}\to{\rm{2(153}}{\rm{.33)gm+1(32)gm}}\\{\rm{338}}{\rm{.66gmBa}}{{\rm{O}}_{\rm{2}}}\to{\rm{306}}{\rm{.66gmBaO+32gm}}{{\rm{O}}_{\rm{2}}}\end{aligned}

For \[{\rm{129}}{\rm{.7gm Ba}}{{\rm{O}}_{\rm{2}}} \to \frac{{{\rm{(129}}{\rm{.7) \times (32)}}}}{{{\rm{338}}{\rm{.66}}}}{\rm{ gm }}{{\rm{O}}_{\rm{2}}}\]

\begin{aligned}\to\frac{{{\rm{(4150}}{\rm{.4)}}}}{{{\rm{338}}{\rm{.66}}}}{\rm{gm}}{{\rm{O}}_{\rm{2}}}\\\to {\rm{12}}{\rm{.25 gm }}{{\rm{O}}_{\rm{2}}}\end{aligned}

Now, it can be obtained –

\begin{aligned}{\rm{32gmof}}{{\rm{O}}_{\rm{2}}}{\rm{=1moleof}}{{\rm{O}}_{\rm{2}}}\\{\rm{12}}{\rm{.25gmof}}{{\rm{O}}_{\rm{2}}}{\rm{=molesof}}{{\rm{O}}_{\rm{2}}}\\{\rm{=}}\frac{{{\rm{(12}}{\rm{.25)\times(1)}}}}{{{\rm{32}}}}{\rm{molesof}}{{\rm{O}}_{\rm{2}}}\\{\rm{=}}\frac{{{\rm{12}}{\rm{.25}}}}{{{\rm{32}}}}{\rm{molesof}}{{\rm{O}}_{\rm{2}}}\\{\rm{=0}}{\rm{.38 moles of }}{{\rm{O}}_{\rm{2}}}\end{aligned}

The formula for volume is PV=NRT.

If, amount of the gas (n) is in moles, temperature (T) is in Kelvin (K), pressure (P) is in atm.

When n=0.3, P=127.4k Pa, T=423K.

When n=0.3, P=127.4k Pa, T=423K.

Converting Pressure kPa into atm –

1 kPa = 0.0098

127.4kPa = (127.4) ×(0.0098)atm

= 1.25atm

Now calculating the volume

\begin{aligned}{\rm{PV=nRT}}\\{\rm{V=}}\frac{{{\rm{nRT}}}}{{\rm{P}}}\\{\rm{=}}\frac{{{\rm{(0}}{\rm{.38)\times(0}}{\rm{.08206)\times(423)}}}}{{{\rm{1}}{\rm{.25}}}}\\{\rm{=}}\frac{{{\rm{13}}{\rm{.2}}}}{{{\rm{1}}{\rm{.25}}}}\\{\rm{=10}}{\rm{.56\;L}}\end{aligned}

Therefore, the value for volume is obtained as V=10.56 L.