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) 1g N₂ ? g NH3 = 61.802 cg N₂ x LR 2 mol NH3 x 17.04 g NH3 = 1 mol N₂ 1 mol NH3 1 x 10² cg N₂ ? g NH3 = 61.802 cg H₂ x How much N₂ remains in the vessel? 1g H₂ 1 x 10² g H₂ X 1g N2₂ 1 x 10² cg N₂ X 1 mol N₂ X 28.02 g N₂ 1 mol H₂ 2.02 g H₂ x 2 mol NH3 x 17.04 g NH3 = 3 mol H₂ 1 mol NH3 You will use the LIMITING REACTANT and determine how much H₂ was USED in the RXN. ? g H₂ USED= 61.802 cg N₂ x 0.75168 g NH3 ******* THEORETICAL YIELD Amount of H₂ Remaining in the Container = H₂ amount given - H₂ amount USED= 3.3756 g NH3 X 1 mol N₂ x 3 mol H₂ x 2.02 g H₂ = 0.13366 g H₂ 28.02 g N₂ 1 mol N₂ 1 mol H₂ 0.61802 g H₂ GIVEN 0.13366 g H₂ USED = 0.48436 g of H2--LEFT OVER = EXCESS
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) 1g N₂ ? g NH3 = 61.802 cg N₂ x LR 2 mol NH3 x 17.04 g NH3 = 1 mol N₂ 1 mol NH3 1 x 10² cg N₂ ? g NH3 = 61.802 cg H₂ x How much N₂ remains in the vessel? 1g H₂ 1 x 10² g H₂ X 1g N2₂ 1 x 10² cg N₂ X 1 mol N₂ X 28.02 g N₂ 1 mol H₂ 2.02 g H₂ x 2 mol NH3 x 17.04 g NH3 = 3 mol H₂ 1 mol NH3 You will use the LIMITING REACTANT and determine how much H₂ was USED in the RXN. ? g H₂ USED= 61.802 cg N₂ x 0.75168 g NH3 ******* THEORETICAL YIELD Amount of H₂ Remaining in the Container = H₂ amount given - H₂ amount USED= 3.3756 g NH3 X 1 mol N₂ x 3 mol H₂ x 2.02 g H₂ = 0.13366 g H₂ 28.02 g N₂ 1 mol N₂ 1 mol H₂ 0.61802 g H₂ GIVEN 0.13366 g H₂ USED = 0.48436 g of H2--LEFT OVER = EXCESS
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) 1g N₂ ? g NH3 = 61.802 cg N₂ x LR 2 mol NH3 x 17.04 g NH3 = 1 mol N₂ 1 mol NH3 1 x 10² cg N₂ ? g NH3 = 61.802 cg H₂ x How much N₂ remains in the vessel? 1g H₂ 1 x 10² g H₂ X 1g N2₂ 1 x 10² cg N₂ X 1 mol N₂ X 28.02 g N₂ 1 mol H₂ 2.02 g H₂ x 2 mol NH3 x 17.04 g NH3 = 3 mol H₂ 1 mol NH3 You will use the LIMITING REACTANT and determine how much H₂ was USED in the RXN. ? g H₂ USED= 61.802 cg N₂ x 0.75168 g NH3 ******* THEORETICAL YIELD Amount of H₂ Remaining in the Container = H₂ amount given - H₂ amount USED= 3.3756 g NH3 X 1 mol N₂ x 3 mol H₂ x 2.02 g H₂ = 0.13366 g H₂ 28.02 g N₂ 1 mol N₂ 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
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 x
1 mol N₂
17.04 g NH3 =
1 mol NH3
1 x 10² cg N₂
? g NH3 = 61.802 cg H₂ x
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
1 mol H₂
2.02 g H₂
How much N2 remains in the vessel?
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
X
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
0.13366 g H₂
NH3
0.61802 g H₂ GIVEN 0.13366 g H2 USED
= 0.48436 g of H₂--LEFT OVER = EXCESS
Definition Definition Chemical reactions involving both oxidation and reduction processes. During a redox reaction, electron transfer takes place in such a way that one chemical compound gets reduced and the other gets oxidized.
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