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We need an ideal gas equation:

\(p \cdot V = n \cdot R \cdot T\)

where p is pressure, V volume, n number of moles, T temperature of gas and R is a gas constant which is \(8.314\frac{J}{{Kmol}}\)

we get,

\(V = \frac{{n \cdot R \cdot T}}{p}\)

Convert pressure from torr to pascal = 101325 Pa

The reaction occurs is,

\(\begin{aligned}{}2C + {O_2} \to 2CO\\F{e_2}{O_3} + 3CO \to 2Fe + 3C{O_2}\\\frac{{n(F{e_2}{O_3})}}{{n(CO)}} &= \frac{1}{3}\\n(CO) = 3.n(F{e_2}{O_3})\end{aligned}\)

Calculating moles of \(F{e_2}{O_3}\)from a simple formula:

\(\begin{aligned}{}n(CO) &= 3.n(F{e_2}{O_3})\\n(F{e_2}{O_3}) &= \frac{{9.07 \cdot {{10}^5}}}{{159.69gmo{l^{ - 1}}}}\\n(F{e_2}{O_3}) &= 5.68 \cdot {10^3}mol\end{aligned}\)

We get,

\(\begin{aligned}{}n(CO) &= 3.n(F{e_2}{O_3})\\n(CO) &= 3 \cdot 5.68 \cdot {10^3}mol\\n(CO) &= 17.04 \cdot {10^3}mol\\\\\frac{{n({O_2})}}{{n(CO)}} &= \frac{1}{2}\\n({O_2}) &= \frac{1}{2}.n(CO)\\n({O_2}) &= \frac{1}{2}{.17.04.10^3}mol\\n({O_2}) &= {8.52.10^3}mol\end{aligned}\)

Volume of oxygen:

\(\begin{aligned}{}V({O_2}) &= \frac{{n.R.T}}{p}\\V({O_2}) &= \frac{{{{8.52.10}^3}mol.8.134\frac{J}{{kmol}}.273.15K}}{{101325Pa}}\\V({O_2}) &= 190.97{m^3} &= 190970d{m^3}\end{aligned}\)

We have 19% of oxygen in the air it means we have 19\(c{m^3}\)of oxygen in 100\(c{m^3}\)of air

\(\begin{aligned}{}V(air) &= \frac{{100c{m^3}(air)}}{{19c{m^3}({O_2})}}{.1.92.10^8}c{m^3}({O_2})\\V(air) &= {1.10^9}c{m^3} &= {3.53.10^4}f{t^3}\\1c{m^3} &= {3.53.10^{ - 5}}f{t^3}\end{aligned}\)

Hence, we need \({3.53.10^{ - 5}}f{t^3}\)of air to convert \(F{e_2}{O_3}\)to iron in a blast furnance.