Stoichiometry Preparation for College Chemistry Luis Avila Columbia

Stoichiometry Preparation for College Chemistry Luis Avila Columbia University Department of Chemistry

The Mole-ratio method Mole-Mole Calculations Mole-Mass Calculations Mass-Mass calculations Yield calculations

Mole-Ratio 2 KMn. O 4 + 6 HCl + 5 H 2 S 1 mol. KMn. O 4 1 mol. KCl 2 mol KMn. O 4 5 mol S 2 KCl +2 Mn. Cl 2 +5 S +8 H 2 O 2 mol KMn. O 4 5 mol H 2 S 8 mol H 2 O How many moles of S can be obtained from 1. 5 mole KMn. O 4 ? 1. 5 mole KMn. O 4 x 5 mole. S 2 mole KMn. O 4 = 3. 8 Mole S

Mole-Mass Calculation 2 KMn. O 4 + 6 HCl + 5 H 2 S 2 KCl +2 Mn. Cl 2 +5 S +8 H 2 O How many g of S can be obtained from 1. 5 g KMn. O 4 ? 1. 5 g KMn. O 4 x 1 mol KMn. O 4 158. 04 g KMn. O 4 x 5 mol S x 32. 07 g S 2 mol KMn. O 4 1 mol S = 80. 9 g S

Limiting Reactant

Limiting Reactant Calculation 16. 8 g 10. 0 g 3 Fe(s) + 4 H 2 O(g) Fe 3 O 4 + 4 H 2 1. Calculate amount of product formed by each reactant 16. 8 g Fe x 10. 0 g H 2 O x 1 mol Fe 55. 85 g Fe x 1 mol Fe 3 O 4 3 mol Fe =. 100 mol Fe 3 O 4 yield of product 1 mol H 2 O 1 mol Fe 3 O 4 x =. 139 mol Fe 3 O 4 18. 02 g H 2 O 4 mol H 2 O 2. The Limiting reactant gives the least amount of product.

Excess Reactant Calculation 16. 8 g 10. 0 g 3 Fe(s) + 4 H 2 O(g) Fe 3 O 4 + 4 H 2 3. Calculate the excess amount by subtracting the reacted amount from the starting quantity 16. 8 g Fe x 1 mol Fe 55. 85 g Fe x 4 mol H 2 O 3 mol Fe x 18. 02 g H 2 O 1 mol H 2 O Excess water: 10. 0 g - 7. 2 g = 2. 7 g unreacted water = 7. 23 g H 2 O

Percentage Yield Calculation Actual Yield Theoretical Yield x 100 In the previous reaction theoretical yield was. 100 mol Fe 3 O 4 x 231. 55 g Fe 3 O 4 1 mol Fe 3 O 4 = 23. 2 g Fe 3 O 4 If the actual amount obtained is 16. 2 g, then the % yield: Percentage Yield = 16. 2 g Fe 3 O 4 23. 2 g Fe 3 O 4 x 100 = 69. 8%