1. Stoichiometry Practice Problem Michelle Lamary

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

3. Write a Complete and Balanced Equation 2H(NO 2 ) + Ba(OH) 2  2H 2 O + Ba(NO 2 ) 2

4. Draw a Column for Each Chemical 2H(NO 2 )+ Ba(OH) 2  2H 2 O +Ba(NO 2 ) 2

5. Write the Amount Given in the Appropriate Column 2H(NO 2 )+ Ba(OH) 2  2H 2 O +Ba(NO 2 ) 2 5.14 g

6. Convert the Given Amount Into Moles 2H(NO 2 )+ Ba(OH) 2  2H 2 O +Ba(NO 2 ) 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

7. Find Moles for Each of the Other Chemicals 2H(NO 2 )+ Ba(OH) 2  2H 2 O +Ba(NO 2 ) 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

8. In Each of the Columns Write the Moles of Given (x) a Fraction 2H(NO 2 )+ Ba(OH) 2  2H 2 O +Ba(NO 2 ) 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

9. The Numerator of the Fraction is the Coefficient of the That Column 2H(NO 2 )+ Ba(OH) 2  2H 2 O +Ba(NO 2 ) 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

10. The Denominator of the Fraction is the Coefficient of the Given Column 2H(NO 2 )+ Ba(OH) 2  2H 2 O +Ba(NO 2 ) 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

11. Do Math and Label as Moles 2H(NO 2 )+ Ba(OH) 2  2H 2 O +Ba(NO 2 ) 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

12. Convert All Moles Into Grams 2H(NO 2 )+ Ba(OH) 2  2H 2 O +Ba(NO 2 ) 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 137.33 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 g 5.14 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 g1.96 g12.5 g

13. Verify the Law of Conservation and Mass 2H(NO 2 )+ Ba(OH) 2  2H 2 O +Ba(NO 2 ) 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 137.33 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 g 5.14 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 g1.96 g12.5 g 14.48 g14.46 g

14. Answer 1.96 grams of water (H 2 O) are produced when 5.14 grams of Hydrogen Nitrite [H(NO 2 )] is reacted with Barium Hydroxide [Ba(OH) 2 ].