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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.

Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Donald J. DeCoste
Chapter9: Chemical Quantities
Section: Chapter Questions
Problem 44QAP: Balance the following chemical equation, and then answer the question below....
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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: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
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.
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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