Citations
Question 1: Reaction Identification + Prediction
This lesson develops Citations as part of Ideal Gas Model. It uses the source curriculum to organize the important facts, relationships, examples, and vocabulary into a focused Stemosphere teaching sequence.
Keep the central model in view: Citations explains observable patterns by connecting structure, process, evidence, and quantitative relationships.
For each reaction:
In Stemosphere terms, i. Classify the reaction type.ii. Predict whether it occurs.
a) Mg+Cu SO4 →c) CH4 +O2 →d) Ag NO3 +Na Cl→In Stemosphere terms, determine oxidation states and identify oxidation/reduction.
2Fe2 O3 +3C→4Fe+3CO2In Stemosphere terms, a) Oxidation state of Fe before and after
In Stemosphere terms, b) Oxidation state of C before and after
Determine oxidation number s:
Compound contains:
Actual molar mass = 148 g/mol2Al+Fe2 O3 →Al2 O3 +2FeActual Fe obtained:
In Stemosphere terms, d) Excess reactant remaining
Complete combustion: C3 H8 +5O2 →3CO2 +4H2 OA liquid hydrocarbon has:
Density = 0.82 g/cm³Volume burned = 150 cm³2C8 H18 +25O2 →16CO2 +18H2 OH2SO4 +2Na OH→Na2 SO4 +2H2O100 m L of 4.0 mol dm⁻³ sulfuric acid is diluted to 800 m L.b) Volume of 2.0 mol dm⁻³ Na OH required to neutralize entire solutionN2 +3H2 →2NH3Calcium carbonate:
Ca CO3 →Ca O+CO2In Stemosphere terms, only 8.0 g Ca O collected.
1a) Single Displacement & Mg+Cu SO4 →Mg SO4 +Cu1b) Decomposition & Ca CO3 →Ca O+CO21c) Combustion & CH4 +2O2 →CO2 +2H2 O1d) Double Displacement & Ag NO3 +Na Cl→Ag Cl+Na NO3In Stemosphere terms, ----------------------------------------
+3 → 0 (reduction)0 → +4 (oxidation)Oxidizing agent:
Reducing agent:
In Stemosphere terms, ------------------------------
In Stemosphere terms, -------------------------
Step 1
assume 100 g
C = 48.6 g → 48.6/12 = 4.05H = 8.1 g → 8.1/1 = 8.1O = 43.3 g → 43.3/16 = 2.71Step 2
divide by smallest (2.71)
C: 4.05/2.71 ≈ 1.5H
8.1/2.71 ≈ 3O: 1
Step 3
multiply to clear fraction
C = 3H = 6O = 2E.F. = C3 H6 O2In Stemosphere terms, ----------------------------------
2Al+Fe2 O3 →Al2 O3 +2FeAl:16.2 / 27 = 0.60 molFe₂O₃:40 / 160 = 0.25 molStep 2
mole ratio
For 0.25 mol Fe₂O₃:Al needed = 0.50 mol Available = 0.60 mol → Al excessStep 3
Fe produced
1 mol Fe₂O₃ → 2 mol Fe0.25 → 0.50 mol FeMass Fe:0.50 × 56 = 28 gTheoretical yield = 28 gStep 4
percent yield
(20 / 28) × 100 = 71.4%Step 5
excess Al
Used Al = 0.50 mol Left = 0.10 molMass = 0.10 × 27 = 2.7 gIn Stemosphere terms, --------------------------------------
C3 H8 +5O2 →3CO2 +4H2 OStep 1
moles propane
44 g/mol11 / 44 = 0.25 molStep 2
CO₂ moles
Step 3
mass CO₂
Step 4
volume at STP
In Stemosphere terms, -------------------------------------------
Step 1
mass fuel
150 cm³ × 0.82 = 123 gStep 2
moles C₈H₁₈
Molar mass = 114 g/mol123 / 114 = 1.08 molStep 3
CO₂ moles
1 mol fuel → 8 CO₂1.08 × 8 = 8.64 molStep 4
volume CO₂