Question

What mass of al is required to completely react with 25.0 grams of mno2?

Answer 1

Mass of MnO2 = 25 g 
The reaction would be 3MnO2 + 4Al --> 3Mn(s) + 2Al2O3
 Molar mass of Al = 26.982 g/mol
 Molar mass of MnO2 = 54.938 + 2(15.999) = 86.936 g/mol
 Calculating the moles = 25 / 86.936 = 0.2876 mol.
 Mole ratio MnO2 and Al considering the equation = 3 mol of MnO : 4 mol of Al
 Calculating the moles of Al = 0.2876 mol MnO2 x (4 mol of Al / 3 mol of MnO)
 Number of moles of Al = 0.3834
 Getting the mass in grams as asked = 0.3834 mol x 26.982 g/mol = 10.34 grams.

Answer 2

10.34 grams Should be the correct answer