Stoichiometry Practice Problem

Michelle Lamary

Question

How many grams of water are produced when 5.14 grams of Hydrogen Nitrite is reacted with Barium Hydroxide?

Write a Complete and Balanced Equation

2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2

Draw a Column for Each Chemical

2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2

Write the Amount Given in the Appropriate Column

2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2

5.14 g

Convert the Given Amount Into Moles

2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2

5.14 g1.0079 g

+14.007 g

+2(15.999) g

47.013 g

5.14 g x 1 mole

1 47.013 g

.109 moles

Find Moles for Each of the Other Chemicals

2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2

5.14 g Moles? Moles? 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

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 ?/? = .109 moles x ?/? = .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

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/? = .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

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 = .109 moles x ½ =

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

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 = .109 moles x ½ =

1.0079 g

+14.007 g

+2(15.999) g

47.013 g

.0545 moles .109 moles .0545 moles

5.14 g x 1 mole

1 47.013 g

.109 moles

Convert All Moles Into Grams

2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2

5.14 g .109 moles x ½ = .109 moles x 1 = .109 moles x ½ =

1.0079 g

+14.007 g

+2(15.999) g

47.013 g

.0545 moles .109 moles .0545 moles137.33 g

+ 2(15.999) g

+ 2(1.0079) g

171.34 g

2(1.0079) g

+ 15.999 g

18.015 g

137.3 g

+ 2(14.007) g

+ 4(15.999) g

229.34 g5.14 g x 1 mole

1 47.013 g .0545 moles x 171.34 g

1 1 mole

.109 moles x 18.015 g

1 1 mole

.0545 moles x 229.34 g

1 1 mole

.109 moles 9.34 g 1.96 g 12.5 g

Verify the Law of Conservation and Mass

2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2

5.14 g .109 moles x ½ = .109 moles x 1 = .109 moles x ½ =

1.0079 g

+14.007 g

+2(15.999) g

47.013 g

.0545 moles .109 moles .0545 moles137.33 g

+ 2(15.999) g

+ 2(1.0079) g

171.34 g

2(1.0079) g

+ 15.999 g

18.015 g

137.3 g

+ 2(14.007) g

+ 4(15.999) g

229.34 g5.14 g x 1 mole

1 47.013 g .0545 moles x 171.34 g

1 1 mole

.109 moles x 18.015 g

1 1 mole

.0545 moles x 229.34 g

1 1 mole

.109 moles 9.34 g 1.96 g 12.5 g

14.48 g 14.46 g

Answer

1.96 grams of water (H2O) are produced when 5.14 grams of Hydrogen Nitrite [H(NO2)] is reacted with Barium Hydroxide [Ba(OH)2].