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When 3.00 g of Mg is ignited in 2.20 g of pure oxygen what is the limiting reactant and what is the theoretical yield of MgO?
Wiki User
∙ 11y ago
Moles Mg = 3.00 g / 24.312 g/mol =0.123
Moles O2 = 2.20 / 32 g/mol = 0.0688
2 Mg + O2 >> 2 MgO
the ratio between Mg and O2 is 2 : 1
0.123 / 2 = 0.0615 moles O2 needed
we have 0.0688 moles of O2 so O2 is in excess and Mg is the limiting reactant
we get 0.123 moles of MgO => 0.123 mol x 40.31 g/mol =4.96 g
Wiki User
∙ 11y ago
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