Silver nitrate reacts with Al to form aluminum nitrate and silver. The balanced chemical reaction is: Al(s) + 3 AgNO, (aq) → AI (NO3)3 (aq) + 3 Ag (s) According to the balanced chemical reaction, 3 moles of AgNO3 reacts with 1 mole of Al to produce 1 mole of aluminum nitrate and 3 moles of Ag. Since, Al is present in excess amount. Thus, silver nitrate is limiting reactant in this reaction. Reaction Type: Single Displacement (Oxidation-Reduction) reaction. 1 Gg = 10-⁹ g The mass of aluminum nitrate produced is calculated as: 1g AgNO, 1x10 Gg AgNO, = 1.495 x 10 ⁹ g 2g AI (NO3)3 = 3.577 Gg AgNO, X I mol AI (NO₂) 3 mol AgNO, The mass of silver produced is calculated as: 1g AgNO, 1x10 Gg AgNO, X 213 g AI (NO₂) X 1 mol Al (NO3)3 2g Ag = 3.577 Gg AgNO, x- 107.87 g Ag 1 mol Ag = 2.271 x 10⁹ g 2g Al = 3.577 Gg AgNO, X 26.98 g Al = 1.89 x 10-10 g I mol Al X 1g AgNO, 1x10 Gg AgNO, The amount of aluminum used in the reaction vessel is calculated as: 1 mol AgNO, 169.87 g AgNO, X X I mol AgNO, 169.87 g AgNO, 3 mol Ag 1 mol AgNO, 169.87 g AgNO, 3 mol AgNO, X X I mol Al 3 mol AgNO3 Thus, the amount of aluminum nitrate present in the container is 1.495 × 10-9 g. The amount of silver present in the container is 2.271 x 10-º g.
Silver nitrate reacts with Al to form aluminum nitrate and silver. The balanced chemical reaction is: Al(s) + 3 AgNO, (aq) → AI (NO3)3 (aq) + 3 Ag (s) According to the balanced chemical reaction, 3 moles of AgNO3 reacts with 1 mole of Al to produce 1 mole of aluminum nitrate and 3 moles of Ag. Since, Al is present in excess amount. Thus, silver nitrate is limiting reactant in this reaction. Reaction Type: Single Displacement (Oxidation-Reduction) reaction. 1 Gg = 10-⁹ g The mass of aluminum nitrate produced is calculated as: 1g AgNO, 1x10 Gg AgNO, = 1.495 x 10 ⁹ g 2g AI (NO3)3 = 3.577 Gg AgNO, X I mol AI (NO₂) 3 mol AgNO, The mass of silver produced is calculated as: 1g AgNO, 1x10 Gg AgNO, X 213 g AI (NO₂) X 1 mol Al (NO3)3 2g Ag = 3.577 Gg AgNO, x- 107.87 g Ag 1 mol Ag = 2.271 x 10⁹ g 2g Al = 3.577 Gg AgNO, X 26.98 g Al = 1.89 x 10-10 g I mol Al X 1g AgNO, 1x10 Gg AgNO, The amount of aluminum used in the reaction vessel is calculated as: 1 mol AgNO, 169.87 g AgNO, X X I mol AgNO, 169.87 g AgNO, 3 mol Ag 1 mol AgNO, 169.87 g AgNO, 3 mol AgNO, X X I mol Al 3 mol AgNO3 Thus, the amount of aluminum nitrate present in the container is 1.495 × 10-9 g. The amount of silver present in the container is 2.271 x 10-º g.
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Transcribed Image Text:14. Balanced Chemical Equation
Reaction Type:
At completion of reactions:
Formula of Reactant A
Formula of Reactant B
Formula of Product C
Formula of Product D
Grams of Reactant A
Grams of Reactant B
Grams of Product C
Grams of Product D
Transcribed Image Text:Silver nitrate reacts with Al to form aluminum nitrate and silver. The balanced chemical reaction is:
Al(s) + 3 AgNO, (aq) AI (NO3)3 (aq) + 3 Ag (s)
According to the balanced chemical reaction, 3 moles of AgNO3 reacts with 1 mole of Al to produce 1
mole of aluminum nitrate and 3 moles of Ag.
Since, Al is present in excess amount. Thus, silver nitrate is limiting reactant in this reaction.
Reaction Type: Single Displacement (Oxidation-Reduction) reaction.
1 Gg = 10⁹ g
The mass of aluminum nitrate produced is calculated as:
1g AgNO3
1x10 Gg AgNO3
= 1.495 x 10-⁹ g
2g AI (NO3)3 = 3.577 Gg AgNO, X
X
1 mol Al (NO₂)
3 mol AgNO,
The mass of silver produced is calculated as:
Ig AgNO,
1x10 Gg AgNO3
X
2g Ag = 3.577 Gg AgNO, x
213 g AI (NO3)3
1 mol Al (NO3)3
107.87 g Ag
1 mol Ag
X
= 2.271 x 10⁹ g
2g Al = 3.577 Gg AgNO, x-
1.89 × 10-¹0 g
X
1 mol AgNO3
169.87 g AgNO,
The amount of aluminum used in the reaction vessel is calculated as:
Ig AgNO3
1x10 Gg AgNO,
1 mol AgNO,
1 mol Al
X
169.87 g AgNO, 3 mol AgNO3
1 mol AgNO3
3 mol Ag
X
169.87 g AgNO, 3 mol AgNO3
26.98 g Al
1 mol Al
Thus, the amount of aluminum nitrate present in the container is 1.495 x 10-⁹ g.
The amount of silver present in the container is 2.271 x 10-⁹ g.
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