Posted by eng on Friday, February 12, 2010 at 2:08pm.
You know it's a limiting reagent problem when both of the starting materials are given. You must determine which is the limiting reagent.
Two calculations will do it.
142 Ba x (2 moles BaO/2 moles Ba) = 142 x 1/1 = 142 BaO molecules formed.
73 O2 x (2 moles BaO/1 moles O2) = 73 x (1/2) = 146 BaO molecules formed.
Both answers can't be correct; the correct one is ALWAYS the smaller. Therefore Ba is the limiting reagent and 142 BaO molecules (formula units) will be formed. Some of the oxygen will remain unreacted. You can determine how much is left easily.
142 Ba atoms x (1 mole O2/2 moles BaO) = 142 x (1/2) = 71 molecules O2 used. We had 73 initially; therefore, 73-71 = 2 molecules O2 remain unreacted.
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