How many grams of Al are required to completely react with 81.2 g of MnO2. What is the mole ratio between MnO2 and Al in the balanced equation? 3MnO2 +4Al--> 3Mn+2Al2O3?

You'd need #"33.6 g"# of aluminium to react with that much manganese dioxide.

So, you've got your balanced chemical equation for this reaction

#3MnO_(2(s)) + 4Al_((s)) -> 3Mn_((s)) + 2Al_2O_(3(s))#

The mole ratio that exists between manganese dioxide and aluminium is simply the ration between the stoichiometric coefficients placed in front of them.

In your case, the balanced chemical equation shows that you have 3 moles of manganese dioxide reacting with 4 moles of aluminium; this means that your mole ratio will be #"3:4"#.

Likewise, the mole ratio between aluminium and manganese dioxide will be #"4:3"# #-># 4 moles of the former need 3 moles of the latter.

You can put this mole ratio to good use by determining how many moles of aluminium are needed to react with the number of moles of manganese dioxide present in 81.2 g. So,

#"81.2 g MnO"_2 * "1 mole"/"86.94 g" = "0.9340 moles"# #MnO_2#

This means that you'll need

#"0.934 moles "MnO_2 * "4 moles Al"/("3 moles "MnO_2) = "1.245 moles Al"#

To determine the number of grams needed to get that many moles use aluminium's molar mass

#"1.245 moles Al" * "26.98 g"/"1 mole Al" = "33.59 g Al"#

Rounded to three sig figs, the answer will be

#m_("aluminium") = "33.6 g"#