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4. Zn (s) + 2 HCI (aq) → ZnCl₂ (aq) + H₂(g) Reaction Type: Single Displacement Grams of Product (List the Product and the Amount): 7.48 x 102 g ZnCl2 and 1.11 x 10¹ g H₂ ? g ZnCl2 = 3.59 x 10² g Zn x 1 mol Zn x 1 mol ZnCl₂ x 65.37 g Zn 1 mol Zn (This is just stoichiometry.) ? g H₂= 3.59 x 10² g Zn x 136.27 g ZnCl₂ = 7.48 x 10² g ZnCl2 1 mol ZnCl2 1 mol Zn x 65.37 g Zn YOU MUST CALCULATE THE AMOUNT OF EXCESS REACTANT LEFT OVER AT THE END OF THE REACTION!!!!!!! 1 mol H₂ x 2.02 g H₂ = 1 mol Zn 1 mol H₂ 1.11 x 10¹ g H₂

4. Zn (s) + 2 HCI (aq) → ZnCl₂ (aq) + H₂(g) Reaction Type: Single Displacement Grams of Product (List the Product and the Amount): 7.48 x 102 g ZnCl2 and 1.11 x 10¹ g H₂ ? g ZnCl2 = 3.59 x 10² g Zn x 1 mol Zn x 1 mol ZnCl₂ x 65.37 g Zn 1 mol Zn (This is just stoichiometry.) ? g H₂= 3.59 x 10² g Zn x 136.27 g ZnCl₂ = 7.48 x 10² g ZnCl2 1 mol ZnCl2 1 mol Zn x 65.37 g Zn YOU MUST CALCULATE THE AMOUNT OF EXCESS REACTANT LEFT OVER AT THE END OF THE REACTION!!!!!!! 1 mol H₂ x 2.02 g H₂ = 1 mol Zn 1 mol H₂ 1.11 x 10¹ g H₂

Chemistry
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
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Calculate the amount of excess reactant left over at the end of the reaction, use the example.

4. Zn (s) + 2 HCI (aq) → ZnCl2 (aq) + H2(g) (This is just stoichiometry.) Reaction Type: Single Displacement Grams of Product (List the Product and the Amount): 7.48 x 10² g ZnCl₂ and 1.11 x 10¹ g H₂ ? g ZnCl₂ = 3.59 x 10² g Zn x ? g H₂= 3.59 x 10² g Zn x 1 mol Zn x 65.37 g Zn 1 mol Zn x 65.37 g Zn mol ZnCl₂ x 1 mol Zn 1 mol H₂ x 1 mol Zn 136.27 g ZnCl₂ = 1 mol ZnCl2 2.02 g H₂ = 1 mol H₂ 7.48 x 10² g ZnCl2 1.11 x 10¹ g H₂ YOU MUST CALCULATE THE AMOUNT OF EXCESS REACTANT LEFT OVER AT THE END OF THE REACTION!!!!!!!
Transcribed Image Text:4. Zn (s) + 2 HCI (aq) → ZnCl2 (aq) + H2(g) (This is just stoichiometry.) Reaction Type: Single Displacement Grams of Product (List the Product and the Amount): 7.48 x 10² g ZnCl₂ and 1.11 x 10¹ g H₂ ? g ZnCl₂ = 3.59 x 10² g Zn x ? g H₂= 3.59 x 10² g Zn x 1 mol Zn x 65.37 g Zn 1 mol Zn x 65.37 g Zn mol ZnCl₂ x 1 mol Zn 1 mol H₂ x 1 mol Zn 136.27 g ZnCl₂ = 1 mol ZnCl2 2.02 g H₂ = 1 mol H₂ 7.48 x 10² g ZnCl2 1.11 x 10¹ g H₂ 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) 1 g N₂ ? g NH3 = 61.802 cg N₂ x LR 2 mol NH3 2 mol NH3 x 1 mol N₂ 17.04 g NH3 1 mol NH3 1 x 10² cg N₂ ? g NH3 = 61.802 cg H₂ x How much N₂ remains in the vessel? 1 g H₂ 1 x 10² g H₂ ? g H₂ USED= 61.802 cg N₂ x 1 g 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 H₂ x 2.02 g H₂ 2 mol NH3 x 17.04 g NH3 = 3 mol H₂ 1 mol NH3 1 mol N₂ x 3 mol H₂ x 2.02 g H₂ 28.02 g N₂ 1 mol N₂ 1 mol H₂ 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
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) 1 g N₂ ? g NH3 = 61.802 cg N₂ x LR 2 mol NH3 2 mol NH3 x 1 mol N₂ 17.04 g NH3 1 mol NH3 1 x 10² cg N₂ ? g NH3 = 61.802 cg H₂ x How much N₂ remains in the vessel? 1 g H₂ 1 x 10² g H₂ ? g H₂ USED= 61.802 cg N₂ x 1 g 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 H₂ x 2.02 g H₂ 2 mol NH3 x 17.04 g NH3 = 3 mol H₂ 1 mol NH3 1 mol N₂ x 3 mol H₂ x 2.02 g H₂ 28.02 g N₂ 1 mol N₂ 1 mol H₂ 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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