Calculating Limiting Reactant.
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Example Limiting Reactant Calculation:A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is the limiting reactant and how much excess reactant remains after the reaction has stopped? How much nitrogen monoxide will be produced in this reaction? 1. First, we need to create a balanced equation for the reaction:4 NH3 + 5 O2 →4 NO + 6 H2O2. Next we can use stoichiometry to calculate how much of one of the reactants will be used in the reaction. Do this by choosing one of the givens and solving for the other reactant. NOTE: It does not matter which reactant you start with. 2.00 g NH3 1 mol NH3 5 mol O2 32.0 g O2 = 4.71 g O2 are needed for this reaction 17.0 g NH3 4 mol NH3 1 mol O2 Because we only have 4.00 g of oxygen but we need 4.71 g oxygen, we know that oxygen is limiting this reaction. Oxygen is the limiting reactant. 3. Next we can calculate how much product will be formed. Use stoichiometry to calculate how much product will be formed starting with the limiting reactant as your given. The limiting reactant controls how much product can be formed. 4.00 g O2 1 mol O2 4 mol NO 30.0 g NO = 3.00 g NO will be formed 32.0 g O2 5 mol O2 1 mol NO 4. Finally, to find the amount of excess reactant, we must calculate how much of the excess reactant (NH3) actually reacted with the limiting reactant (oxygen). 4.00 g O2 1 mol O2 4 mol NH3 17.0 g NH3 = 1.70 g NH3 32.0 g O2 5 mol O2 1 mol NH3We're not finished yet though. 1.70 g is the amount of ammonia that reacted, not what is left over. To find the amount of excess reactant remaining, subtract the amount that reacted from the amount in the original sample. Started with used up have left excess NH3 = 0.30 g 2.00 g NH3 - 1.70 g NH3 = 0.30 g NH3