Respuesta :
Answer: The mass of iron (III) oxide needed to react is 67.4 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 carbon = 19.0 g
Molar mass of carbon = 12 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of carbon}=\frac{19.0g}{12g/mol}=1.58mol[/tex]
The chemical equation for the reaction of iron (III) oxide and carbon follows:
[tex]2Fe_2O_3+3C\rightarrow 3CO_2+4Fe[/tex]
By Stoichiometry of the reaction:
3 moles of carbon reacts with 2 moles of iron (III) oxide
So, 1.58 moles of carbon will react with = [tex]\frac{2}{3}\times 1.58=0.422mol[/tex] of iron (III) oxide
Now, calculating the mass of iron (III) oxide by using equation 1, we get:
Molar mass of iron (III) oxide = 159.70 g/mol
Moles of iron (III) oxide = 0.422 moles
Putting values in equation 1, we get:
[tex]0.422mol=\frac{\text{Mass of iron (III) oxide}}{159.70g/mol}\\\\\text{Mass of iron (III) oxide}=(0.422mol\times 159.70g/mol)=67.4g[/tex]
Hence, the mass of iron (III) oxide needed to react is 67.4 g
