Respuesta :
Answer : The mass of iron(III)sulfide is, 5.4288 g
Solution : Given,
Mass of iron, Fe = 3 g
Mass of sulfur, [tex]S_8[/tex] = 2.5 g
Molar mass of Fe = 56 g/mole
Molar mass of [tex]S_8[/tex] = 256 g/mole
Molar mass of iron(III)sulfide, [tex]Fe_2S_3[/tex] = 208 g/mole
- The balanced chemical reaction is,
[tex]16Fe(s)+3S_8(s)\rightarrow 8Fe_2S_3(s)[/tex]
First we have to calculate the moles of iron and sulfur.
[tex]\text{ Moles of Fe}=\frac{\text{ Mass of Fe}}{\text{ Molar mass of Fe}}=\frac{3g}{56g/mole}=0.054moles[/tex]
[tex]\text{ Moles of }S_8=\frac{\text{ Mass of }S_8}{\text{ Molar mass of }S_8}=\frac{2.5g}{256g/mole}=0.0098moles[/tex]
- From the balanced reaction, we conclude that
16 moles of Fe react with 3 moles of [tex]S_8[/tex]
0.054 moles of Fe react with [tex]\frac{3}{16}\times 0.054=0.010125[/tex] moles of [tex]S_8[/tex]
Therefore, the excess reagent in this reaction is, Fe and limiting reagent is, [tex]S_8[/tex]
Now we have to calculate the moles of FeS.
As, 3 moles of [tex]S_8[/tex] gives 8 moles of [tex]Fe_2S_3[/tex]
So, 0.0098 moles of [tex]S_8[/tex] gives [tex]\frac{8}{3}\times 0.0098=0.0261[/tex] moles of [tex]F_2eS_3[/tex]
The moles of [tex]Fe_2S_3[/tex] = 0.0261 moles
Now we have to calculate the mass of [tex]Fe_2S_3[/tex].
Mass of [tex]Fe_2S_3[/tex] = Moles of [tex]Fe_2S_3[/tex] × Molar mass of [tex]Fe_2S_3[/tex]
Mass of [tex]Fe_2S_3[/tex] = 0.0261 g × 208 g/mole = 5.4288 g
Therefore, the mass of iron(III)sulfide is, 5.4288 g
