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Write the balanced equation for the following situation. List the reaction type. Tell the amounts of every substance that remains in the container at the end of the reaction. Assume that all reactions go to completion If only stoichiometry, tell how much of the excess reactant is used!!!! Reaction Type: a. Combination Reaction b. Decomposition Reaction c. Single Displacement / THIS IS ONE TYPE OF Oxidation Reduction Reaction d. Precipitation Reaction e. Gaseous Reaction f. Neutralization Reaction g. Combustion Reaction Follow the format used in the image. 1.95 g iron(III) chloride and 25.00 mL of a 0.250 M sodium hydroxide solution 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 N₂ LR x 1 mol N₂ x 28.02 g N₂ 1 x 10² cg N₂ ? g NH3 = 61.802 cg H₂ x 1g H₂ 1 x 10² g H₂ 1 x 10² cg N₂ X Amount of H₂ Remaining in the Container 1 mol H₂ x 2.02 g H₂ 2 mol NH₂ x 17.04 g NH3 0.75168 g NH3 ******* THEORETICAL YIELD 1 mol N₂ 1 mol NH How much N₂ remains in the vessel? You will use the LIMITING REACTANT and determine how much H₂ was USED in the RXN. ? g H₂ USED= 61.802 cg N2 x 1g N₂ x 1 mol N₂ x 3 mol H₂ 28.02 g N₂ 1 mol N₂ H₂ amount given - H₂ amount USED 2 mol NH x 17.04 g NH3 = 3.3756 g NH3 3 mol H₂ 1 mol Nha x 2.02 g H₂= 0.13366 g H₂ 1 mol H₂ 0.61802 g H₂ GIVEN 0.13366 g H₂ USED -0.48436 g of H₂--LEFT OVER-EXCESS

Write the balanced equation for the following situation. List the reaction type. Tell the amounts of every substance that remains in the container at the end of the reaction. Assume that all reactions go to completion If only stoichiometry, tell how much of the excess reactant is used!!!! Reaction Type: a. Combination Reaction b. Decomposition Reaction c. Single Displacement / THIS IS ONE TYPE OF Oxidation Reduction Reaction d. Precipitation Reaction e. Gaseous Reaction f. Neutralization Reaction g. Combustion Reaction Follow the format used in the image. 1.95 g iron(III) chloride and 25.00 mL of a 0.250 M sodium hydroxide solution 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 N₂ LR x 1 mol N₂ x 28.02 g N₂ 1 x 10² cg N₂ ? g NH3 = 61.802 cg H₂ x 1g H₂ 1 x 10² g H₂ 1 x 10² cg N₂ X Amount of H₂ Remaining in the Container 1 mol H₂ x 2.02 g H₂ 2 mol NH₂ x 17.04 g NH3 0.75168 g NH3 ******* THEORETICAL YIELD 1 mol N₂ 1 mol NH How much N₂ remains in the vessel? You will use the LIMITING REACTANT and determine how much H₂ was USED in the RXN. ? g H₂ USED= 61.802 cg N2 x 1g N₂ x 1 mol N₂ x 3 mol H₂ 28.02 g N₂ 1 mol N₂ H₂ amount given - H₂ amount USED 2 mol NH x 17.04 g NH3 = 3.3756 g NH3 3 mol H₂ 1 mol Nha x 2.02 g H₂= 0.13366 g H₂ 1 mol H₂ 0.61802 g H₂ GIVEN 0.13366 g H₂ USED -0.48436 g of H₂--LEFT OVER-EXCESS

Living by Chemistry
Living by Chemistry
2nd Edition
ISBN:9781464142314
Author:Angelica M. Stacy
Publisher:Angelica M. Stacy
ChapterU4: Toxins: Stoichiometry, Solution Chemistry, And Acids And Bases
Section: Chapter Questions
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Write the balanced equation for the following situation. List the reaction type. Tell the amounts of every substance that remains in the container at the end of the reaction. Assume that all reactions go to completion. If only stoichiometry, tell how much of the excess reactant is used!!!! Reaction Type: a. Combination Reaction b. Decomposition Reaction c. Single Displacement / THIS IS ONE TYPE OF Oxidation Reduction Reaction d. Precipitation Reaction e. Gaseous Reaction f. Neutralization Reaction g. Combustion Reaction Follow the format used in the image. 1.95 g iron(III) chloride and 25.00 mL of a 0.250 M sodium hydroxide solution 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 N₂ LR 1 x 10² cg N₂ ? g NH3 = 61.802 cg H₂ x 1g H₂ 1 x 10² g H₂ x x 1 mol N₂ x 28.02 g N₂ 1 mol H₂ x 2.02 g H₂ 2 mol NH3 x 17.04 g NH3 = 0.75168 g NH3 ******* THEORETICAL YIELD 1 mol N₂ 1 mol NH3 2 mol NH3 x 17.04 g NH3 = 3.3756 g NH3 3 mol H₂ 1 mol NH3 How much N₂ remains in the vessel? You will use the LIMITING REACTANT and determine how much H₂ was USED in the RXN. ? g H₂ USED = 61.802 cg N₂ x 1g N₂ x 1 mol N₂ x 3 mol H₂ 1 x 10² cg N₂ 28.02 g N₂ 1 mol N₂ Amount of H₂ Remaining in the Container = H₂ amount given H₂ amount USED= = x 2.02 g H₂ = 0.13366 g H₂ 1 mol H₂ 0.61802 g H₂ GIVEN 0.13366 g H₂ USED 0.48436 g of H₂--LEFT OVER = EXCESS
Transcribed Image Text:Write the balanced equation for the following situation. List the reaction type. Tell the amounts of every substance that remains in the container at the end of the reaction. Assume that all reactions go to completion. If only stoichiometry, tell how much of the excess reactant is used!!!! Reaction Type: a. Combination Reaction b. Decomposition Reaction c. Single Displacement / THIS IS ONE TYPE OF Oxidation Reduction Reaction d. Precipitation Reaction e. Gaseous Reaction f. Neutralization Reaction g. Combustion Reaction Follow the format used in the image. 1.95 g iron(III) chloride and 25.00 mL of a 0.250 M sodium hydroxide solution 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 N₂ LR 1 x 10² cg N₂ ? g NH3 = 61.802 cg H₂ x 1g H₂ 1 x 10² g H₂ x x 1 mol N₂ x 28.02 g N₂ 1 mol H₂ x 2.02 g H₂ 2 mol NH3 x 17.04 g NH3 = 0.75168 g NH3 ******* THEORETICAL YIELD 1 mol N₂ 1 mol NH3 2 mol NH3 x 17.04 g NH3 = 3.3756 g NH3 3 mol H₂ 1 mol NH3 How much N₂ remains in the vessel? You will use the LIMITING REACTANT and determine how much H₂ was USED in the RXN. ? g H₂ USED = 61.802 cg N₂ x 1g N₂ x 1 mol N₂ x 3 mol H₂ 1 x 10² cg N₂ 28.02 g N₂ 1 mol N₂ Amount of H₂ Remaining in the Container = H₂ amount given H₂ amount USED= = x 2.02 g H₂ = 0.13366 g H₂ 1 mol H₂ 0.61802 g H₂ GIVEN 0.13366 g H₂ USED 0.48436 g of H₂--LEFT OVER = EXCESS
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