Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter3: Chemical Reactions
Section: Chapter Questions
Problem 68GQ: Balance the following equations: (a) for the reaction to produce "superphosphate" fertilizer...
Related questions
Question
What is the amount of excess reactant left over at the end of the reaction. Use the first image as an example.
![3. Ca(OH)2 (aq) + H₂SO4 (aq) → + 2 H₂O (1) + CaSO4 (aq) (This is LIMITING REACTANT: H₂SO4 is the Limiting Reactant)
Reaction Type: Neutralization
Grams of Product (List the Product and the Amount): 0.180 g H₂O and 0.681 g CaSO4
? g H₂O
=
1.839 g Ca(OH)2 x
? g H₂O = 25.0 mL H₂SO4 X
1 mol Ca(OH)2 X
74.10 g Ca(OH)2
2 mol H₂O x 18.02 g H₂O
1 mol Ca(OH)2 1 mol H₂O
= 0.8944 g H₂O
1 L H₂SO4 x 0.200 mol H₂SO4 x 2 mol H₂O x 18.02 g H₂O = 0.180 g H₂O
1000 mL H₂SO4 1 L H2SO4 1 mol H₂SO4 1 mol H₂O
ONLY USE THE Limiting Reactant TO DETERMINE THE AMOUNT OF THE REMAINING PRODUCT(S).
? g H₂O = 25.0 mL H₂SO4 x
1 L H₂SO4 x 0.200 mol H₂SO4 x 1 mol CaSO4 x 136.14 g CaSO4 = 0.681 g CaSO4
1000 mL H₂SO4 1 L H2SO4 1 mol H₂SO4 1 mol CaSO4
YOU MUST CALCULATE THE AMOUNT OF EXCESS REACTANT LEFT OVER AT THE END OF THE REACTION!!!!!!!](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff43825b3-7357-4f9c-824e-d428b47940ce%2Fa50ca30d-695e-4b89-bf94-000877dbfa3a%2Fda2994_processed.jpeg&w=3840&q=75)
Transcribed Image Text:3. Ca(OH)2 (aq) + H₂SO4 (aq) → + 2 H₂O (1) + CaSO4 (aq) (This is LIMITING REACTANT: H₂SO4 is the Limiting Reactant)
Reaction Type: Neutralization
Grams of Product (List the Product and the Amount): 0.180 g H₂O and 0.681 g CaSO4
? g H₂O
=
1.839 g Ca(OH)2 x
? g H₂O = 25.0 mL H₂SO4 X
1 mol Ca(OH)2 X
74.10 g Ca(OH)2
2 mol H₂O x 18.02 g H₂O
1 mol Ca(OH)2 1 mol H₂O
= 0.8944 g H₂O
1 L H₂SO4 x 0.200 mol H₂SO4 x 2 mol H₂O x 18.02 g H₂O = 0.180 g H₂O
1000 mL H₂SO4 1 L H2SO4 1 mol H₂SO4 1 mol H₂O
ONLY USE THE Limiting Reactant TO DETERMINE THE AMOUNT OF THE REMAINING PRODUCT(S).
? g H₂O = 25.0 mL H₂SO4 x
1 L H₂SO4 x 0.200 mol H₂SO4 x 1 mol CaSO4 x 136.14 g CaSO4 = 0.681 g CaSO4
1000 mL H₂SO4 1 L H2SO4 1 mol H₂SO4 1 mol CaSO4
YOU MUST CALCULATE THE AMOUNT OF EXCESS REACTANT LEFT OVER AT THE END OF THE REACTION!!!!!!!
![2.
N₂ (g) + 3 H₂(g) → 2 NH3 (g) (This is LIMITING REACTANT: N₂ is the Limiting Reactant)
Grams of Product (List the Product and the Amount): 0.75168 g NH3
Reaction Type: Combination or Single Displacement (which is also called Oxidation-Reduction)
? g NH3 = 61.802 cg N₂ x 1g №₂
2 mol NH3 x 17.04 g NH3
1 mol N₂
1 mol NH3
LR
? g NH3 = 61.802 cg H₂ x
How much N₂ remains in the vessel?
1 x 10² cg N₂
? g H₂ USED = 61.802 cg N₂ x
1 g H₂
1 x 10² g H₂
1g N₂
1 x 10² cg N₂
X
X
X
1 mol N₂ x
28.02 g N₂
You will use the LIMITING REACTANT and determine how much H2 was USED in the RXN.
1 mol N₂ x 3 mol H₂ x 2.02 g H₂ =
28.02 g N₂ 1 mol N₂ 1 mol H₂
1 mol H₂ x
2.02 g H₂
2 mol NH3 x 17.04 g NH3 =
3 mol H₂
1 mol NH3
Amount of H₂ Remaining in the Container = H₂ amount given H₂ amount USED =
=
0.75168 g NH3 ******* THEORETICAL YIELD
3.3756 g NH3
0.13366 g H₂
0.61802 g H₂ GIVEN - 0.13366 g H₂ USED
0.48436 g of H2--LEFT OVER = EXCESS](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff43825b3-7357-4f9c-824e-d428b47940ce%2Fa50ca30d-695e-4b89-bf94-000877dbfa3a%2Fwq5se2q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:2.
N₂ (g) + 3 H₂(g) → 2 NH3 (g) (This is LIMITING REACTANT: N₂ is the Limiting Reactant)
Grams of Product (List the Product and the Amount): 0.75168 g NH3
Reaction Type: Combination or Single Displacement (which is also called Oxidation-Reduction)
? g NH3 = 61.802 cg N₂ x 1g №₂
2 mol NH3 x 17.04 g NH3
1 mol N₂
1 mol NH3
LR
? g NH3 = 61.802 cg H₂ x
How much N₂ remains in the vessel?
1 x 10² cg N₂
? g H₂ USED = 61.802 cg N₂ x
1 g H₂
1 x 10² g H₂
1g N₂
1 x 10² cg N₂
X
X
X
1 mol N₂ x
28.02 g N₂
You will use the LIMITING REACTANT and determine how much H2 was USED in the RXN.
1 mol N₂ x 3 mol H₂ x 2.02 g H₂ =
28.02 g N₂ 1 mol N₂ 1 mol H₂
1 mol H₂ x
2.02 g H₂
2 mol NH3 x 17.04 g NH3 =
3 mol H₂
1 mol NH3
Amount of H₂ Remaining in the Container = H₂ amount given H₂ amount USED =
=
0.75168 g NH3 ******* THEORETICAL YIELD
3.3756 g NH3
0.13366 g H₂
0.61802 g H₂ GIVEN - 0.13366 g H₂ USED
0.48436 g of H2--LEFT OVER = EXCESS
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