Answer: 48.4 g of [tex]Cl_2[/tex] will be produced from 25.5 g of iron
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Fe=\frac{25.5 g}{56g/mol}=0.455moles[/tex]
The balanced chemical reaction is:
[tex]2Fe(s)+3Cl_2(g)\rightarrow 2FeCl_3(s)[/tex]
According to stoichiometry :
2 moles of [tex]Fe[/tex] require = 3 moles of [tex]Cl_2[/tex]
Thus 0.455 moles of [tex]Fe[/tex] will require=[tex]\frac{3}{2}\times 0.455=0.682moles[/tex] of [tex]Cl_2[/tex]
Mass of [tex]Cl_2=moles\times {\text {Molar mass}}=0.682moles\times 71g/mol=48.4g[/tex]
Thus 48.4 g of [tex]Cl_2[/tex] will be produced from 25.5 g of iron