Follow the format used in the image. 5.92 g of sodium oxalate is reacted with 5.92 of calcium chloride

Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter3: Chemical Reactions
Section: Chapter Questions
Problem 70QRT
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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.

5.92 g of sodium oxalate is reacted with 5.92 of calcium chloride

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
2 mol NH3 x 17.04 g NH3 =
1 mol N₂
1 mol NH3
LR
? g NH3 = 61.802 cg H₂ x
How much N₂ remains in the vessel?
1 x 10² cg N₂
1 g H₂
1 x 10² g H₂
X
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.
=
0.75168 g NH3 *****
0.75168 g NH3
=
3.3756 g NH3
? g H₂ USED = 61.802 cg N₂ x 1g N₂
x 1 mol N₂ x 3 mol H₂ x 2.02 g H₂
1 x 10² cg N₂ 28.02 g N₂ 1 mol N₂ 1 mol H₂
Amount of H₂ Remaining in the Container = H₂ amount given - H₂ amount USED= 0.61802 g H₂ GIVEN - 0.13366 g H₂ USED
0.48436 g of H2--LEFT OVER = EXCESS
******* THEORETICAL YIELD
0.13366 g H₂
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) ? g NH3 = 61.802 cg N₂ x 1g N₂ 2 mol NH3 x 17.04 g NH3 = 1 mol N₂ 1 mol NH3 LR ? g NH3 = 61.802 cg H₂ x How much N₂ remains in the vessel? 1 x 10² cg N₂ 1 g H₂ 1 x 10² g H₂ X 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. = 0.75168 g NH3 ***** 0.75168 g NH3 = 3.3756 g NH3 ? g H₂ USED = 61.802 cg N₂ x 1g N₂ x 1 mol N₂ x 3 mol H₂ x 2.02 g H₂ 1 x 10² cg N₂ 28.02 g N₂ 1 mol N₂ 1 mol H₂ Amount of H₂ Remaining in the Container = H₂ amount given - H₂ amount USED= 0.61802 g H₂ GIVEN - 0.13366 g H₂ USED 0.48436 g of H2--LEFT OVER = EXCESS ******* THEORETICAL YIELD 0.13366 g H₂
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