Burning 12.0 g of Mg produces 18.1 g of MgO. For 2Mg + O2 -> 2MgO, find the percent yield.

The mole ratio Mg:MgO is 1:1.
moles Mg = 12.0 g Mg / 24.31 g/mol = 0.494 mol Mg
moles MgO = 0.494 mol MgO
theoretical mass MgO = 0.494 mol * 40.31 g/mol = 19.9 g MgO
percent yield = (actual yield / theoretical yield) * 100%
percent yield = (18.1 g / 19.9 g) * 100% = 91.0%.