Manganese(IV) oxide reacts with aluminum to form elemental manganese and aluminum oxide: 3MnO2+4Al→3Mn+2Al2O3 What mass of...
Manganese(IV) oxide reacts with aluminum to form elemental manganese and aluminum oxide: 3MnO2+4Al→3Mn+2Al2O3 What mass of Al is required to completely react with 30.0 g MnO2?
Homework Answers
3MnO2+4Al→3Mn+2Al2O3
m = 30 g of MnO2
MW of MnO2 = 86.9368
mol = 30/86.9368 = 0.345
3 mol of MnO2 : 4 mol of Al
therefore we need
4/3*0.345 = 0.46 mol of Al
MW of Al = 26.9
mass = mol*MW = 0.46*26.9 = 12.37 grams of Al
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