Stoichiometry Practice Problems Whole Topic Review

© Mr. D. Scott; CHS2 3 Hg 2 I HCl + 14 HNO 3  6 HgCl HIO NO + 10 H 2 O Use the balanced equation to determine the following:  Identity of the excess & limiting reactants –Mass of un-reacted excess when the reaction is complete –Mass of iodic acid produced if there is a 97.0% yield –Volume of NO produced at 35.0°C and 1.25 atm (100% Yield) –Density of NO gas when cooled to 12.0°C at atm Data: 21.0 g Hg 2 I g HCl Excess NHO 3 Excess-Limiting compare Limiting = HAVE < NEED Limiting Excess HAVENEED Convert from grams to moles Use the mole ratio Convert from moles back to grams Solution:

© Mr. D. Scott; CHS3 3 Hg 2 I HCl + 14 HNO 3  6 HgCl HIO NO + 10 H 2 O Use the balanced equation to determine the following: –Identity of the excess & limiting reactants  Mass of un-reacted excess when the reaction is complete –Mass of iodic acid produced if there is a 97.0% yield –Volume of NO produced at 35.0°C and 1.25 atm (100% Yield) –Density of NO gas when cooled to 12.0°C at atm Data: 21.0 g Hg 2 I g HCl Excess NHO 3 Mass of un-reacted excess Solution: Excess Reactant5.00 g HCl Used amount g HCl Un-reacted excess 0.32 g HCl From previous calculation Left over when reaction is complete

© Mr. D. Scott; CHS4 3 Hg 2 I HCl + 14 HNO 3  6 HgCl HIO NO + 10 H 2 O Use the balanced equation to determine the following: –Identity of the excess & limiting reactants –Mass of un-reacted excess when the reaction is complete  Mass of iodic acid produced if there is a 97.0% yield –Volume of NO produced at 35.0°C and 1.25 atm (100% Yield) –Density of NO gas when cooled to 12.0°C at atm Data: 21.0 g Hg 2 I g HCl Excess NHO 3 Mass of iodic acid at 97.0% yield Solution: Start with the limiting reactant Use the mole ratio from the balanced equation Convert from moles to grams. This would give the 100% stoichiometric yield if we stopped here in the calculation.` Since we have a 97.0% yield, we must reduce the theoretical by multiplying by the percent. This is the actual amount of product formed at 97.0% yield.

© Mr. D. Scott; CHS5 3 Hg 2 I HCl + 14 HNO 3  6 HgCl HIO NO + 10 H 2 O Use the balanced equation to determine the following : –Identity of the excess & limiting reactants –Mass of un-reacted excess when the reaction is complete –Mass of iodic acid produced if there is a 97.0% yield  Volume of NO produced at 35.0°C and 1.25 atm (100% Yield) –Density of NO gas when cooled to 12.0°C at atm Data: 21.0 g Hg 2 I g HCl Excess NHO 3 Volume of NO Solution: Start with the limiting reactant Use the mole ratio from the balanced equation Transfer this value into the ideal gas equation. And, substitute the rest of the gas variables.

© Mr. D. Scott; CHS6 3 Hg 2 I HCl + 14 HNO 3  6 HgCl HIO NO + 10 H 2 O Use the balanced equation to determine the following : –Identity of the excess & limiting reactants –Mass of un-reacted excess when the reaction is complete –Mass of iodic acid produced if there is a 97.0% yield –Volume of NO produced at 35.0°C and 1.25 atm (100% Yield)  Density of NO gas when cooled to 12.0°C at atm Data: 21.0 g Hg 2 I g HCl Excess NHO 3 Density of NO Solution:

© Mr. D. Scott; CHS7 2 KMnO HCl  2 KCl + 2 MnCl Cl H 2 O Use the balanced equation to determine the following:  Identity of the excess & limiting reactants –Mass of un-reacted excess when the reaction is complete –Mass of KCl produced if there is a 93.5% yield –Volume of Cl 2 produced at 48.0°C and atm ( 100% Yield ) –Density of Cl 2 gas when cooled to 5.00°C at 1.78 atm Data: 4.75 g KMnO g HCl Excess-Limiting compare Limiting = HAVE < NEED Limiting Excess HAVENEED Convert from grams to moles Use the mole ratio Convert from moles back to grams Solution:

© Mr. D. Scott; CHS8 2 KMnO HCl  2 KCl + 2 MnCl Cl H 2 O Use the balanced equation to determine the following: –Identity of the excess & limiting reactants  Mass of un-reacted excess when the reaction is complete –Mass of KCl produced if there is a 93.5% yield –Volume of Cl 2 produced at 48.0°C and atm ( 100% Yield ) –Density of Cl 2 gas when cooled to 5.00°C at 1.78 atm Mass of un-reacted excess Solution: Excess Reactant4.75 g KMnO 4 Used amount g KMnO 4 Un-reacted excess 0.13 g KMnO 4 From previous calculation Left over when reaction is complete Data: 4.75 g KMnO g HCl

© Mr. D. Scott; CHS9 Convert from moles to grams. This would give the 100% stoichiometric yield if we stopped here in the calculation.` 2 KMnO HCl  2 KCl + 2 MnCl Cl H 2 O Use the balanced equation to determine the following: –Identity of the excess & limiting reactants –Mass of un-reacted excess when the reaction is complete  Mass of KCl produced if there is a 93.5% yield –Volume of Cl 2 produced at 48.0°C and atm ( 100% Yield ) –Density of Cl 2 gas when cooled to 5.00°C at 1.78 atm Mass of KCl at 93.5% yield Solution: Start with the limiting reactant Use the mole ratio from the balanced equation Since we have a 93.5% yield, we must reduce the theoretical by multiplying by the percent. This is the actual amount of product formed at 93.5% yield. Data: 4.75 g KMnO g HCl

© Mr. D. Scott; CHS10 2 KMnO HCl  2 KCl + 2 MnCl Cl H 2 O Use the balanced equation to determine the following : –Identity of the excess & limiting reactants –Mass of un-reacted excess when the reaction is complete –Mass of KCl produced if there is a 93.5% yield  Volume of Cl 2 produced at 48.0°C and atm (100% Yield) –Density of Cl 2 gas when cooled to 5.00°C at 1.78 atm Volume of Cl 2 Solution: Start with the limiting reactant Use the mole ratio from the balanced equation Transfer this value into the ideal gas equation. And, substitute the rest of the gas variables. Data: 4.75 g KMnO g HCl

© Mr. D. Scott; CHS11 2 KMnO HCl  2 KCl + 2 MnCl Cl H 2 O Use the balanced equation to determine the following : –Identity of the excess & limiting reactants –Mass of un-reacted excess when the reaction is complete –Mass of KCl produced if there is a 93.5% yield –Volume of Cl 2 produced at 48.0°C and atm ( 100% Yield )  Density of Cl 2 gas when cooled to 5.00°C at 1.78 atm Density of Cl 2 Solution: Data: 4.75 g KMnO g HCl