How many Oxygen atoms are in the compound Ca(NO3)2
Classify:
2H₂ + O₂ → 2H₂O
Synthesis
N₂ + 3H₂ → 2NH₃
If you have 3 moles of H₂, how many moles of NH₃ can be produced?
2 Moles NH3
Balance:
H₂ + O₂ → H₂O
2H₂ + O₂ → 2H₂O
2H₂ + O₂ → 2H₂O
If you have 2 moles of H₂ and 2 moles of O₂, which is limiting?
You only need 1 mol O₂ for 2 mol H₂, but you have 2 mol O₂ → H₂ is limiting
What is the molar mass of H2SO4? (include units of molar mass)
What is 98.09 g/mole?
Classify:
CaCO₃ → CaO + CO₂
Decomposition
2H₂ + O₂ → 2H₂O
If you have 4 moles of H₂, how many moles of H₂O form?
4 Moles H2O
Balance:
Fe + O₂ → Fe₂O₃
4Fe + 3O₂ → 2 Fe₂O₃
N₂ + 3H₂ → 2NH₃
If you have 1 mole N₂ and 2 moles H₂, which is limiting?
Need 3 mol H₂ for 1 mol N₂, but only have 2 → H₂ is limiting
How many grams are in 2.50 moles of NaCl?
What is 146 grams NaCl ?
Classify:
Zn + 2HCl → ZnCl₂ + H₂
Single Replacement
N₂ + 3H₂ → 2NH₃
How many moles of NH₃ are produced from 25.0 grams of N₂?
1.78 Moles NH3
Balance:
C₂H8 + O₂ → CO₂ + H₂O
C₂H8 + 4O₂ → 2CO₂ + 4H₂O
2H₂ + O₂ → 2H₂O
If you have 4 moles H₂ and 1 mole O₂, which is limiting?
Need 2 mol H₂ for 1 mol O₂, but have extra H₂ → O₂ is limiting
How many grams are in 3.0 × 10²³ molecules of NH₃?
8.5 Grams NH3
Predict the products, and classify:
C₃H₈ + O₂ →
CO2 and H2O; Combustion
2H₂ + O₂ → 2H₂O
How many grams of H₂O are produced from 67.0 grams of O₂?
75.5 grams H2O
Balance:
Al + HCl → AlCl₃ + H₂
2Al + 6HCl → 2AlCl₃ + 3H₂
2H₂ + O₂ → 2H₂O
If you start with 4 moles H₂ and 3 moles O₂, how many moles of H₂O form?
4 mol H₂O produced
How many atoms of oxygen are in 2.25 moles of Ca(NO₃)₂?
8.127x1023 Oxygen Atoms
Predict the products, balance, and classify:
Na₂SO₄ + BaCl₂ →
Na2SO4 + BaCl2 → BaSO4 + 2NaCl
Double Replacement
When silver nitrate reacts with sodium chloride, solid silver chloride forms along with aqueous sodium nitrate.
Balanced equation:
AgNO₃ + NaCl → AgCl + NaNO₃
Molar masses:
AgNO₃ ≈ 169.9 g/mol
AgCl ≈ 143.3 g/mol
25.0 g AgNO₃ → 0.147 mol AgNO₃
Mole ratio 1:1 → 0.147 mol AgCl
→ 21.1 g AgCl
Balance:
C₄H₁₀ + O₂ → CO₂ + H₂O
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
N₂ + 3H₂ → 2NH₃
If you start with 28 g of N₂ and 6.0 g of H₂:
Need 3 mol H₂ per 1 mol N₂ → perfectly matched
→ Neither is limiting (perfect ratio)
Now find product:
1 mol N₂ → 2 mol NH₃
2 mol NH₃ × 17.0 g/mol = 34.0 g NH₃