If 33 moles of MnO2 combine with 40 moles of Al, how many moles of Mn can form? What is the limiting reagent? 3MnO2(s)+4Al(s)⟶3Mn(s)+2Al2O3(s)

Answer 1

Just use the mole ratios from the balanced reaction:

(33 mol MnO₂) * (3 mol Mn / 3 mol MnO₂) = 33 moles of Mn.

(40 mol Al) * (3 mol Mn / 4 mol Al) = 30 moles of Mn.

So the answer is 30 moles of Mn are formed, with Al being the limiting reactant and 3 moles of Mn left over.