Answer: The mass of ferrous sulfide required is 0.615 g
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)
Given mass of iron (III) oxide = 0.56 g
Molar mass of iron (III) oxide = 159.7 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of iron (III) oxide}=\frac{0.56g}{159.7g/mol}=0.0035mol[/tex]
For the given chemical reaction:
[tex]4FeS+7O_2\rightarrow 2Fe_2O_3+4SO_2[/tex]
By Stoichiometry of the reaction:
2 moles of iron (III) oxide is produced from 4 moles of ferrous sulfide.
So, 0.0035 moles of iron (III) oxide will be produced from = [tex]\frac{4}{2}\times 0.0035=0.007mol[/tex] of ferrous sulfide.
Now, calculating the mass of ferrous sulfide from equation 1, we get:
Molar mass of ferrous sulfide = 87.92 g/mol
Moles of ferrous sulfide = 0.007 moles
Putting values in equation 1, we get:
[tex]0.007mol=\frac{\text{Mass of ferrous sulfide}}{87.92g/mol}\\\\\text{Mass of ferrous sulfide}=0.615g[/tex]
Hence, the mass of ferrous sulfide required is 0.615 g
