For the reaction, calculate how many grams of the product form when 2.8 g of Mg completely reacts. Assume that there is more than enough of the other reactant.
2Mg(s)+O2(g)-->2MgO(s)

4.64g MgO(s)

Explanation:

Given 2Mg(s) + O₂(g) => 2MgO(s)

           2.8g     excess        (?)

  => 2.8g/24.31 g/mol

       = 0.115 mol. Mg(s)  => 0.115 mol. MgO(s)

grams MgO(s) produced  = 40.31g/mol. x 0.115 mol. = 4.64g (Theoritical Yield)

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