Answer: The mass of aluminium required is 1.38 grams.
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 [tex]Mn_3O_4[/tex] = 4.40 g
Molar mass of [tex]Mn_3O_4[/tex] = 229 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of }Mn_3O_4=\frac{4.40g}{228.8g/mol}=0.0192mol[/tex]
For the given chemical equation:
[tex]3Mn_3O_4+8Al\rightarrow 9Mn+4Al_2O_3[/tex]
By Stoichiometry of the reaction:
3 moles of [tex]Mn_3O_4[/tex] reacts with 8 moles of aluminium
So, 0.0192 moles of [tex]Mn_3O_4[/tex] will react with = [tex]\frac{8}{3}\times 0.0192=0.0512mol[/tex] of aluminium
Now, calculating the mass of aluminium by using equation 1, we get:
Molar mass of aluminium = 27 g/mol
Moles of aluminium = 0.0512 moles
Putting values in equation 1, we get:
[tex]0.0512mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.0512mol\times 27g/mol)=1.38g[/tex]
Hence, the mass of aluminium required is 1.38 grams.