Answer: 0.000187 mg of iron is produced.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Mg=\frac{0.120g}{24g/mol}=0.005moles[/tex]
[tex]3Mg+2FeCl_2\rightarrow 2Fe+3MgCl_2[/tex]
As [tex]FeCl_2[/tex] is the excess reagent, [tex]Mg[/tex] is the limiting reagent.
According to stoichiometry :
3 moles of [tex]Mg[/tex] give = 2 moles of [tex]Fe[/tex]
Thus 0.005 moles of [tex]Mg[/tex] give =[tex]\frac{2}{3}\times 0.005=0.0033moles[/tex] of [tex]Fe[/tex]
Mass of [tex]Fe=moles\times {\text {Molar mass}}=0.0033moles\times 56g/mol=0.187g=0.000187mg[/tex] (1g=1000mg)
Thus 0.000187 mg of iron is produced.