What was the actual amount of magnesium oxide (Mgo) produced if the percent yield was 81.7% when 62.7 g of magnesium metal (Mg) reacted with an excess amount of carbon dioxide (CO) ?
2 Mg + CO2 - 2 MgO + c
(Molar mass (in g/mol) of: Mg = 24.3 C = 12.01: 0 - 16.0)
1) Find the number of moles of Mg in 62.7gm by dividing by the molecular weight.
2) For every mole of Mg, you get one mole of MgO (x 81.7%)
3) Multiply the number of moles of MgO by the molecular weight of MgO
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