Iron rusts according to the following reaction: $4Fe (s) + 3O_2 (g) + 4H_2O -> 2Fe_2O_3(H_2O)_2$. If 1.4 liters of oxygen were added to excess water and iron at a pressure of 0.97 atms and a temperature of 55 C, how many grams of iron would rust?

Md Ariful Islam

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Md Ariful Islam

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Let us readdress it....iron metal is oxidized to $Fe_2O_3$...

$2Fe(s) +3H_2O(l) rarr Fe_2O_3+6H^+ +6e^(-)$ $(i)$

Dioxygen is reduced to water...

$1/2O_2 + 2H^+ +2e^(-) rarr H_2O(l)$ $(ii)$

And $1xx(i)+3xx(ii)$ gives....

$2Fe(s) +cancel(3H_2O(l)) +3/2O_2 + cancel(6H^+) +cancel(6e^(-)) rarr Fe_2O_3+cancel(6H^+) +cancel(3H_2O(l))+cancel(6e^(-))$

…. and we cancel to give...

$2Fe(s) +3/2O_2 rarr Fe_2O_3$

And note that this equation assumes COMPLETE oxidation....iron oxide chemistry, i.e. rust chemistry is a very broad church...and many iron oxides and hydrous oxides are known. And so we calculate the moles of dioxygen gas....and plug this value back into the given equation....

$n_"dioxygen"=(PV)/(RT)=(0.97*atmxx1.4*L)/(0.0821*(L*atm)/(K*mol)*328.15*K)~=0.05*mol$....and given the equation...$4/3*"equiv"$ of iron metal will oxidize...approx. $4*g$..

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