Burning magnesium in the presence of oxygen forms magnesium oxide. Write the balanced equation. If 10.2 g of Mg reacts with 10.5 g of the oxygen gas, and 11.9 g of the product is formed, what is the limiting reactant and percent yield?
Burning magnesium in the presence of oxygen forms magnesium oxide. Write the balanced equation. If 10.2 g of Mg reacts with 10.5 g of the oxygen gas, and 11.9 g of the product is formed, what is the limiting reactant and percent yield?
Given, mass of Mg reacted = 10.2 g
mass of O2 reacted = 10.5 g
mass of product MgO formed, actual yield = 11.9 g
thus, moles of each reactant can be calculated as,
moles of Mg = = = 0.420 mol
and, moles of O2 = = = 0.328 mol
Now, the balanced equation for the above reaction can be expressed as,
2 Mg (s) + O2 (g) ----> 2MgO (s)
Clearly, from the above reaction,
2 mol of Mg react with = 1 mol of O2
so, 0.420 mol of Mg react with = mol of O2
= 0.210 mol of O2
Thus, moles of O2 present in the reaction is in excess, hence the reaction is limited by moles of Mg available.
So, the limiting reagent is Mg.
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