Answer: 10.3 g of Al is required to completely react with 25.0 g [tex]MnO_2[/tex]
Explanation:
The balanced chemical equation for the reaction is:
[tex]3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3[/tex]
To find the moles we use the formula= [tex]\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
moles of [tex]MnO_2=\frac{25.0}{87g/mol}=0.287moles[/tex]
According to stoichiometry:
3 moles of [tex]MnO_2[/tex] reacts with = 4 moles of aluminium
0.287 moles of [tex]MnO_2[/tex] reacts with =[tex]\frac{4}{3}\times 0.287=0.383[/tex] moles of aluminium
Mass of aluminium =[tex]moles\times {\text {Molar Mass}}=0.383mol\times 27g/mol=10.3g[/tex]
Thus 10.3 g of Al is required to completely react with 25.0 g [tex]MnO_2[/tex]
