Answer: 2
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Fe_3O_4=\frac{275g}{233.5g/mol}=1.18moles[/tex]
[tex]3Fe(s)+4H_2O(g)\rightarrow Fe_3O_4(s)+4H_2(g)[/tex]
According to stoichiometry :
1 mole of [tex]Fe_3O_4[/tex] are produced by = 4 moles of [tex]H_2O[/tex]
Thus 1.18 moles of [tex]Fe_3O_4[/tex] will be produced by=[tex]\frac{4}{1}\times 1.18=4.72moles[/tex] of [tex]H_2O[/tex]
Mass of [tex]H_2O=moles\times {\text {Molar mass}}=4.72moles\times 18g/mol=85.0g[/tex]
Thus 85.0 g of [tex]H_2O[/tex] will be required and 2 steps are required to get the answer.