What is the minimum quantity of Iron required to react with steam to produce the same quantity of hydrogen gas as produced by the reaction of 3 gm of Al with dil. NaOH?

$\text{No}\text{. of Mol of Al =}\frac{\text{Given mass (in gra)}}{\text{g - atomic mass}}=\frac{3}{27}=\frac{1}{9}mole$ 

$2Al+2NaOH+2H_{2}O\to 2NaAl{O}_2+3H_2\uparrow$

$\because 2molAl=3mol{H}_2$

$\therefore \frac{3}{2}\times \frac{1}{9}=\frac{1}{6}mol{H}_2$

$Now:3Fe+\underset{\text{(steam)}}{4H_{2}O}\to F{e}_3{O}_4+4H_2\uparrow$

$\because 4mol{H}_2=3molFe$

$\therefore \frac{1}{6}mol{H}_2=\frac{3}{4}\times \frac{1}{6}=\frac{1}{8}molFe$

Mass (in gram) = Moles × g-atomic mass

$=\frac{1}{8}\times 56=7gFe$