the balanced equation for the following situation. List the reaction type. Write 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. 13.33 cg of barium chloride is reacted with 135.2 mL of a 0.250 M sulfuric acid 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 NH₂ 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 1 mol N₂ x 2 mol NH₁ x 17.04 g NH₁ = 0.75168 g NH; ******* THEORETICAL YIELD 28.02 g N₂ 1 mol N₂ 1 mol NH x 1 mol H₂ x 2.02 g H₂ 2 mol NH x 17.04 g NH₁ = 3.3756 g NH₁ 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 H: 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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the balanced equation for the following situation. List the reaction type.
Write
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.
13.33 cg of barium chloride is reacted with 135.2 mL of a 0.250 M sulfuric acid
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 NH₂
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
1 x 10² cg N₂ 28.02 g N₂
? g NH3 = 61.802 cg H₂ x 1g H₂
1 x 10² g H₂
x
1 mol H₂ x
2.02 g H₂
2 mol NH₁ x 17.04 g NH₁ = 0.75168 g NH; ******* THEORETICAL YIELD
1 mol N₂ 1 mol NH
2 mol NH x 17.04 g NH₁ = 3.3756 g NH₁
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 H: 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:the balanced equation for the following situation. List the reaction type. Write 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. 13.33 cg of barium chloride is reacted with 135.2 mL of a 0.250 M sulfuric acid 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 NH₂ 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 1 x 10² cg N₂ 28.02 g N₂ ? g NH3 = 61.802 cg H₂ x 1g H₂ 1 x 10² g H₂ x 1 mol H₂ x 2.02 g H₂ 2 mol NH₁ x 17.04 g NH₁ = 0.75168 g NH; ******* THEORETICAL YIELD 1 mol N₂ 1 mol NH 2 mol NH x 17.04 g NH₁ = 3.3756 g NH₁ 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 H: 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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