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Limiting/Excess Reagent Practice Problem
Michelle Lamary
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Question Using the following completed and balanced reaction:
2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2 Determine which chemical is the limiting reagent, which is the excess reagent, and how much excess there is if 5.14 grams of H(NO2) is reacted with 9.02 grams of Ba(OH)2
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Draw a Column for Each Chemical and for Each Equation
2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2
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Write the Amounts Given in the Appropriate Columns
2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2 5.14 g 9.99 g
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Convert the Given Amounts Into Moles
2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2 5.14 g g g +2(15.999) g g 5.14 g x 1 mole g .109 moles 9.99 g 137.33 2(1.0079) 2(15.999) g 9.99 g x 1 mole g .0583 moles
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Find Moles for Each of the Other Chemicals
2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2 5.14 g Moles? g g +2(15.999) g g 5.14 g x 1 mole g .109 moles 9.99 g 137.33 2(1.0079) 2(15.999) g 9.99 g x 1 mole g .0583 moles
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In Each of the Columns Write the Moles of Given (x) a Fraction
2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2 5.14 g .109 moles x ?/? = g g +2(15.999) g g 5.14 g x 1 mole g .109 moles .0583 moles x ?/? = 9.99 g 137.33 2(1.0079) 2(15.999) g 9.99 g x 1 mole g .0583 moles
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The Numerator of the Fraction is the Coefficient of the That Column
2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2 5.14 g .109 moles x 1/? = .109 moles x 2/? = g g +2(15.999) g g 5.14 g x 1 mole g .109 moles .0583 moles x 2/? = 9.99 g 137.33 2(1.0079) 2(15.999) g 9.99 g x 1 mole g .0583 moles .0583 moles x 1/? =
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The Denominator of the Fraction is the Coefficient of the Given Column
2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2 5.14 g .109 moles x ½ = .109 moles x 2/2 = g g +2(15.999) g g 5.14 g x 1 mole g .109 moles .0583 moles x 2/1 = 9.99 g 137.33 2(1.0079) 2(15.999) g 9.99 g x 1 mole g .0583 moles .0583 moles x 1/1 =
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Do Math and Label As Moles
2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2 5.14 g .109 moles x ½ = .109 moles x 1 = g g +2(15.999) g g 5.14 g x 1 mole g .109 moles .0545 moles .0583 moles x 2 = .117 moles 9.99 g 137.33 2(1.0079) 2(15.999) g 9.99 g x 1 mole g .0583 moles .0583 moles x 1 =
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2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2
Determine which is the Limiting (small) and Excess (large) and cross out the Excess 2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2 5.14 g .109 moles x ½ = .109 moles x 1 = g g +2(15.999) g g 5.14 g x 1 mole g .109 moles .0545 moles .0583 moles x 2 = .117 moles 9.99 g 137.33 2(1.0079) 2(15.999) g 9.99 g x 1 mole g .0583 moles .0583 moles x 1 =
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Convert All Moles to Grams
2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2 5.14 g .109 moles x ½ = .109 moles x 1 = g g +2(15.999) g g 5.14 g x 1 mole g .109 moles .0545 moles g + 2(15.999) g + 2(1.0079) g g .0545 moles x g mole 9.34 g 2(1.0079) g g g 137.3 g + 2(14.007) g + 4(15.999) g g .0583 moles x 2 = .117 moles 9.99 g 137.33 2(1.0079) 2(15.999) .109 moles x g mole .0545 moles x g mole 1.96 g 12.5 g 9.99 g x 1 mole g .0583 moles .0583 moles x 1 =
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Verify Law of Conservation of Mass
2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2 5.14 g .109 moles x ½ = .109 moles x 1 = g g +2(15.999) g g 5.14 g x 1 mole g .109 moles 14.48 g .0545 moles g + 2(15.999) g + 2(1.0079) g g .0545 moles x g mole 9.34 g 2(1.0079) g g g 137.3 g + 2(14.007) g + 4(15.999) g g .0583 moles x 2 = .117 moles 9.99 g 137.33 2(1.0079) 2(15.999) .109 moles x g mole .0545 moles x g mole 1.96 g 14.46 g 12.5 g 9.99 g x 1 mole g .0583 moles .0583 moles x 1 =
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Answer Limiting Reagent: H(NO2) Excess Reagent: Ba(OH)2
Amount of Excess: = .65 g
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