Answer: 77.47 g of [tex]Fe_2O_3[/tex] will be produced from 100.0 g of Iron (III) hydroxide
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Fe(OH)_3=\frac{100.0g}{106.9g/mol}=0.935moles[/tex]
The balanced chemical equation is:
[tex]2Fe(OH)_3(s)\rightarrow Fe_2O_3(s)+3H_2O(g)[/tex]
According to stoichiometry :
2 moles of [tex]Fe(OH)_3[/tex] produce = 1 mole of [tex]Fe_2O_3[/tex]
Thus 0.935 moles of [tex]Fe(OH)_3[/tex] will producee=[tex]\frac{1}{2}\times 0.935=0.468moles[/tex] of [tex]Fe_2O_3[/tex]
Mass of [tex]Fe_2O_3=moles\times {\text {Molar mass}}=0.468moles\times 159.7g/mol=74.74g[/tex]
Thus 77.47 g of [tex]Fe_2O_3[/tex] will be produced from 100.0 g of Iron (III) hydroxide
