Answer: 7.98 grams of [tex]Fe_2O_3[/tex] are produced if 10.7 grams of [tex]Fe(OH)_3[/tex] are reacted.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Putting values in equation 1, we get:
[tex]\text{Moles of} Fe(OH)_3=\frac{10.7g}{106.87g/mol}=0.100mol[/tex]
The chemical equation for the reaction is
[tex]2Fe(OH)_3\rightarrow Fe_2O_3+3H_2O[/tex]
By Stoichiometry of the reaction:
2 moles of [tex]Fe(OH)_3[/tex] produce = 1 mole of [tex]Fe_2O_3[/tex]
So, 0.100 moles of [tex]Fe(OH)_3[/tex] produce= [tex]\frac{1}{2}\times 0.100=0.05mol[/tex] of [tex]Fe_2O_3[/tex]
Mass of [tex]Fe_2O_3[/tex] =[tex]moles\times {\text{Molar Mass}}=0.05mol\times 159.69g/mol=7.98g[/tex]
Hence 7.98 grams of [tex]Fe_2O_3[/tex] are produced if 10.7 grams of [tex]Fe(OH)_3[/tex] are reacted.
