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

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.

10.34 grams Should be the correct answer