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`\ttcolor{#FF9494}{m}\ttcolor{#FFD1D1}{k}\ttcolor{#FFE3E1}{t}\ttcolor{#FFF5E4}{u'}` 

`a)`

`n_(Fe) = (16,8)/(56) = 0,3`  `(mol)`

$\begin{array}{l}4Fe + 3{O_2} \xrightarrow{t^o} 2F{e_2}{O_3}\end{array}$

`n_(O_2) = 3/4xx0,3 = 0,225`  `(mol)`

`V_(O_2) = 0,225xx22,4 = 5,04`  `(g)` 

$\begin{array}{l} b)\\ F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\ {n_{F{e_2}{O_3}}} = \dfrac{{0,3 \times 2}}{4} = 0,15\, (mol)\\ {n_{{H_2}S{O_4}}} = 3{n_{F{e_2}{O_3}}} = 0,45\, (mol)\\ {m_{{H_2}S{O_4}}} = 0,45 \times 98 = 44,1 (g) \end{array}$