MassMass HW 7 2 Grams 1 Moles 2

Mass-Mass HW 7 -2 Grams 1 → Moles 2 → Grams 2

Problem: COEFFICIENTS of the balanced chemical equation represent MOLE RATIOS but in the lab the only instrument we have for measuring amounts of atoms are scales. 30. 0 g C 1 mole C ( 12. 0 g C ) = 2. 50 moles Solution: Mass-Mass Calculations require you to link the mole concept idea of counting by weighing with the mole ratio concept.

How many grams of Fe 2 O 3 could be produced from the reaction of 167. 4 g of Fe? Coefficients Moles of Fe PT Moles of Fe 2 O 3 4 Fe + 3 O 2 → 2 Fe 2 O 3 167. 4 g Fe PT ? g Step #1: g of Fe → moles of Fe PT = Periodic Table Step #2: moles of Fe → moles of Fe 2 O 3 Step #3: moles of Fe 2 O 3 → grams of Fe 2 O 3

How many grams of Fe 2 O 3 could be produced from the reaction of 167. 4 g of Fe? Moles of Fe PT 4 Fe + 3 O 2 → 2 Fe 2 O 3 167. 4 g Fe ? g PT = Periodic Table 167. 4 g Fe ( 1 mole Fe 55. 85 g Fe ) = 2. 998 moles Fe

How many grams of Fe 2 O 3 could be produced from the reaction of 167. 4 g of Fe? 2. 998 moles Fe Coefficients Moles of Fe 2 O 3 Moles of Fe 4 Fe + 3 O 2 → 2 Fe 2 O 3 167. 4 g Fe ? g PT = Periodic Table 2. 998 moles Fe ( 2 mole Fe 2 O 3 4 moles Fe ) = 1. 500 moles Fe 2 O 3

How many grams of Fe 2 O 3 could be produced from the reaction of 167. 4 g of Fe? 0. 9999 moles Fe 2 O 3 Moles of Fe 2 O 3 4 Fe + 3 O 2 → 2 Fe 2 O 3 167. 4 g Fe PT ? g PT = Periodic Table 1. 500 moles Fe 2 O 3 159. 7 g Fe 2 O 3 ( 1 moles Fe O ) = 239. 6 g Fe O 2 3

How many grams of Fe 2 O 3 could be produced from the reaction of 167. 4 g of Fe? Coefficients Moles of Fe 2 O 3 Moles of Fe PT 4 Fe + 3 O 2 → 2 Fe 2 O 3 167. 4 g Fe PT ? g PT = Periodic Table 167. 4 g Fe 1 mole Fe ( 55. 85 g Fe )( 2 mole Fe 2 O 3 4 moles Fe = 239. 6 g Fe 2 O 3 ) ( 1 moles Fe O) 159. 7 g Fe 2 O 3 2 3

1 A) Calculate the number grams of Al needed to prepare 153 g of Al 2 O 3. Coefficients Moles of Al 2 O 3 Moles of Al PT 4 Al + 3 O 2 → 2 Al 2 O 3 ? g Al PT 153 g PT = Periodic Table 153 g Al 2 O 3 ( 101. 96 g Al O )( 1 mole Al 2 O 3 2 3 4 mole Al 2 moles Al 2 O 3 = 81. 0 g Al )( 26. 98 g Al 1 mole Al )

1 B) Calculate the number grams of O 2 needed to prepare 153 g of Al 2 O 3. Coefficients Moles of Al 2 O 3 Moles of O 2 PT 4 Al + 3 O 2 → 2 Al 2 O 3 ? g O 2 PT 153 g PT = Periodic Table 153 g Al 2 O 3 ( 101. 96 g Al O )( 1 mole Al 2 O 3 2 3 3 mole O 2 2 moles Al 2 O 3 = 72. 0 g Al )( ) 32. 00 g O 2 1 mole Al

N 2 H 4 + 3 O 2 → 2 NO 2 + 2 H 2 O 64. 0 g ? g HW 7 -2 #2 A 64. 0 g N 2 H 4 2 mole NO 2 (32. 0 g N H )( 1 mole N 2 H 4 2 4 = 184 g NO 2 2 4 ) 46. 0 g NO 2 1 moles NO 2

N 2 H 4 + 3 O 2 → 2 NO 2 + 2 H 2 O 64. 0 g ? g HW 7 -1 #2 B 64. 0 g N 2 H 4 2 mole H 2 O (32. 0 g N H )( 1 mole N H ) ( 1 mole N 2 H 4 2 4 = 72. 0 g H 2 O 2 4 ) 18. 0 g H 2 O 1 moles H 2 O

C 2 H 6 O + 3 O 2 → 2 CO 2 + 3 H 2 O 50. 0 g ? g HW 7 -2 : #3 A 50. 0 g C 2 H 6 O ( 1 mole C 2 H 6 O 46. 0 g C 2 H 6 O )( 2 mole CO 2 1 mole C 2 H 6 O = 95. 7 g CO 2 )( 44. 0 g CO 2 1 mole CO 2 )

C 2 H 6 O + 3 O 2 → 2 CO 2 + 3 H 2 O 92. 0 g ? g HW 7 -2 : #3 B 92. 0 g C 2 H 6 O ( 1 mole C 2 H 6 O 46. 0 g C 2 H 6 O )( 3 mole H 2 O 1 mole C 2 H 6 O = 108 g H 2 O )( 18. 0 g H 2 O 1 mole H 2 O )