Limiting/Excess Reagent Practice Problem Michelle Lamary

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

Draw a Column for Each Chemical and for Each Equation 2H(NO2) + Ba(OH)2  2H2O + Ba(NO2)2

Write the Amounts Given in the Appropriate Columns 2H(NO2) + Ba(OH)2  2H2O + Ba(NO2)2 5.14 g 9.99 g

Convert the Given Amounts Into Moles 2H(NO2) + Ba(OH)2  2H2O + Ba(NO2)2 5.14 g 1.0079 g +14.007 g +2(15.999) g 47.013 g 5.14 g x 1 mole 1 47.013 g .109 moles 9.99 g 137.33 2(1.0079) 2(15.999) 171.34 g 9.99 g x 1 mole 1 171.34 g .0583 moles

Find Moles for Each of the Other Chemicals 2H(NO2) + Ba(OH)2  2H2O + Ba(NO2)2 5.14 g Moles? 1.0079 g +14.007 g +2(15.999) g 47.013 g 5.14 g x 1 mole 1 47.013 g .109 moles 9.99 g 137.33 2(1.0079) 2(15.999) 171.34 g 9.99 g x 1 mole 1 171.34 g .0583 moles

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 ?/? = 1.0079 g +14.007 g +2(15.999) g 47.013 g 5.14 g x 1 mole 1 47.013 g .109 moles .0583 moles x ?/? = 9.99 g 137.33 2(1.0079) 2(15.999) 171.34 g 9.99 g x 1 mole 1 171.34 g .0583 moles

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/? = 1.0079 g +14.007 g +2(15.999) g 47.013 g 5.14 g x 1 mole 1 47.013 g .109 moles .0583 moles x 2/? = 9.99 g 137.33 2(1.0079) 2(15.999) 171.34 g 9.99 g x 1 mole 1 171.34 g .0583 moles .0583 moles x 1/? =

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 = 1.0079 g +14.007 g +2(15.999) g 47.013 g 5.14 g x 1 mole 1 47.013 g .109 moles .0583 moles x 2/1 = 9.99 g 137.33 2(1.0079) 2(15.999) 171.34 g 9.99 g x 1 mole 1 171.34 g .0583 moles .0583 moles x 1/1 =

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 = 1.0079 g +14.007 g +2(15.999) g 47.013 g 5.14 g x 1 mole 1 47.013 g .109 moles .0545 moles .0583 moles x 2 = .117 moles 9.99 g 137.33 2(1.0079) 2(15.999) 171.34 g 9.99 g x 1 mole 1 171.34 g .0583 moles .0583 moles x 1 =

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 = 1.0079 g +14.007 g +2(15.999) g 47.013 g 5.14 g x 1 mole 1 47.013 g .109 moles .0545 moles .0583 moles x 2 = .117 moles 9.99 g 137.33 2(1.0079) 2(15.999) 171.34 g 9.99 g x 1 mole 1 171.34 g .0583 moles .0583 moles x 1 =

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 = 1.0079 g +14.007 g +2(15.999) g 47.013 g 5.14 g x 1 mole 1 47.013 g .109 moles .0545 moles 137.33 g + 2(15.999) g + 2(1.0079) g 171.34 g .0545 moles x 171.34 g 1 1 mole 9.34 g 2(1.0079) g + 15.999 g 18.015 g 137.3 g + 2(14.007) g + 4(15.999) g 229.34 g .0583 moles x 2 = .117 moles 9.99 g 137.33 2(1.0079) 2(15.999) .109 moles x 18.015 g 1 1 mole .0545 moles x 229.34 g 1 1 mole 1.96 g 12.5 g 9.99 g x 1 mole 1 171.34 g .0583 moles .0583 moles x 1 =

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 = 1.0079 g +14.007 g +2(15.999) g 47.013 g 5.14 g x 1 mole 1 47.013 g .109 moles 14.48 g .0545 moles 137.33 g + 2(15.999) g + 2(1.0079) g 171.34 g .0545 moles x 171.34 g 1 1 mole 9.34 g 2(1.0079) g + 15.999 g 18.015 g 137.3 g + 2(14.007) g + 4(15.999) g 229.34 g .0583 moles x 2 = .117 moles 9.99 g 137.33 2(1.0079) 2(15.999) .109 moles x 18.015 g 1 1 mole .0545 moles x 229.34 g 1 1 mole 1.96 g 14.46 g 12.5 g 9.99 g x 1 mole 1 171.34 g .0583 moles .0583 moles x 1 =

Answer Limiting Reagent: H(NO2) Excess Reagent: Ba(OH)2 Amount of Excess: 9.99-9.34 = .65 g