To solve this problem we need to know first how many mols of water we have. To do that we use the Ideal gas law with the that provided:
[tex]\begin{gathered} PV=nRT \\ n=\frac{PV}{RT}=\frac{1atm\cdot50L}{0.082\frac{amt\cdot L}{K\cdot mol}\cdot523.15K}=1.166mol \end{gathered}[/tex]according to the chemical equation we know that 3 mols of water react with 2 mols of iron, we can use this information as a conversion factor to know how many mols of iron react with 1.166 mols of water:
[tex]1.166mols\text{ of water}\cdot\frac{2mols\text{ of Fe}}{3\text{ mols of water}}=0.777mols\text{ of Fe}[/tex]and last we have to convert those mols into mass. To do that we use the molar mass of Iron 55.845g/mol:
[tex]m_{Fe}=n_{Fe}\cdot Mm_{Fe}=0.777mols\cdot55.845g/mol=43.939\text{ g}[/tex]the mas of Iron is 43.939g
