Reacting Masses
Section: 3. Stoichiometry | Syllabus: Cambridge AS Level Physics 9702
What are Reacting Masses? Reacting masses are calculations that use the balanced equation and mole ratios to work out the masses of reactants needed or products formed in a chemical reaction. These calculations combine: Balanced chemical equations (for mole ratios) Relative formula masses (Mr values) The mole formula (mass = moles × Mr) The Key Principle The balanced equation tells us the mole ratio in which substances react: Example: 2H₂ + O₂ → 2H₂O This tells us: 2 moles of H₂ react with 1 mole of O₂ to make 2 moles of H₂O The mole ratio is 2:1:2 If we double everything, 4 moles H₂ react with 2 moles O₂ to make 4 moles H₂O If we halve everything, 1 mole H₂ reacts with 0.5 moles O₂ to make 1 mole H₂O Step-by-Step Method for Reacting Mass Calculations Step 1: Write the balanced equation Step 2: Calculate Mr for each substance needed Step 3: Convert the given mass to moles (moles = mass ÷ Mr) Step 4: Use the mole ratio from the equation to find moles of unknown substance Step 5: Convert moles back to mass (mass = moles × Mr) Remember: Mass → Moles → Use ratio → Moles → Mass Example 1: Basic Reacting Mass Calculation Question: What mass of magnesium oxide is produced when 12 g of magnesium burns completely in oxygen?
2Mg + O₂ → 2MgO (Ar: Mg = 24, O = 16) Step 1: Equation is already balanced ✓ Step 2: Calculate Mr values Mr of Mg = 24 Mr of MgO = 24 + 16 = 40 Step 3: Convert given mass to moles Moles of Mg = 12 ÷ 24 = 0.5 mol Step 4: Use mole ratio From equation: 2 moles Mg produces 2 moles MgO Ratio is 2:2 or 1:1 So 0.5 moles Mg produces 0.5 moles MgO Step 5: Convert back to mass Mass of MgO = 0.5 × 40 = 20 g Answer: 20 g of magnesium oxide is produced.
Example 2: Different Mole Ratios Question: What mass of ammonia can be made from 28 g of nitrogen? N₂ + 3H₂ → 2NH₃ (Ar: N = 14, H = 1) Step 1: Equation is balanced ✓ Step 2: Calculate Mr values Mr of N₂ = 2 × 14 = 28 Mr of NH₃ = 14 + (3 × 1) = 17 Step 3: Convert given mass to moles Moles of N₂ = 28 ÷ 28 = 1 mol Step 4: Use mole ratio From equation: 1 mole N₂ produces 2 moles NH₃ Ratio is 1:2 So 1 mole N₂ produces 2 moles NH₃ Step 5: Convert back to mass Mass of NH₃ = 2 × 17 = 34 g Answer: 34 g of ammonia can be made.
Example 3: Finding a Reactant Mass Question: What mass of oxygen is needed to completely react with 8 g of methane? CH₄ + 2O₂ → CO₂ + 2H₂O (Ar: C = 12, H = 1, O = 16) Step 1: Equation is balanced ✓ Step 2: Calculate Mr values Mr of CH₄ = 12 + (4 × 1) = 16 Mr of O₂ = 2 × 16 = 32 Step 3: Convert given mass to moles Moles of CH₄ = 8 ÷ 16 = 0.5 mol Step 4: Use mole ratio From equation: 1 mole CH₄ needs 2 moles O₂ Ratio is 1:2 So 0.5 moles CH₄ needs 1 mole O₂ Step 5: Convert back to mass Mass of O₂ = 1 × 32 = 32 g Answer: 32 g of oxygen is needed.
Example 4: More Complex Ratios Question: What mass of iron is produced when 80 g of iron oxide reacts with excess carbon? 2Fe₂O₃ + 3C → 4Fe + 3CO₂ (Ar: Fe = 56, O = 16, C = 12) Step 1: Equation is balanced ✓ Step 2: Calculate Mr values Mr of Fe₂O₃ = (2 × 56) + (3 × 16) = 112 + 48 = 160 Mr of Fe = 56 Step 3: Convert given mass to moles Moles of Fe₂O₃ = 80 ÷ 160 = 0.5 mol Step 4: Use mole ratio From equation: 2 moles Fe₂O₃ produces 4 moles Fe Ratio is 2:4 or 1:2 So 0.5 moles Fe₂O₃ produces 1 mole Fe Step 5: Convert back to mass Mass of Fe = 1 × 56 = 56 g Answer: 56 g of iron is produced.
Working with Decimal Moles Example 5: What mass of carbon dioxide is produced when 6 g of carbon burns completely? C + O₂ → CO₂ (Ar: C = 12, O = 16) Steps: Mr of C = 12, Mr of CO₂ = 44 Moles of C = 6 ÷ 12 = 0.5 mol Ratio: 1 mole C produces 1 mole CO₂ (1:1) So 0.5 moles C produces 0.5 moles CO₂ Mass of CO₂ = 0.5 × 44 = 22 g Answer: 22 g of carbon dioxide.
Using Mole Ratio Table Method An alternative method is to use a table: Example: 2Mg + O₂ → 2MgO with 12 g of Mg 2Mg + O₂ → 2MgO Mole ratio 2 1 2 Mr 24 32 40 Mass (given) 12 g - ? Moles 0.5 0.25 0.5 Mass (answer) 12 g 8 g 20 g The "Excess" Keyword When a question says "with excess..." it means there's more than enough of that reactant, so you don't need to worry about it: Example: "What mass of copper is produced when 16 g of copper oxide reacts with excess carbon?" The word "excess" tells you: There's plenty of carbon available All the copper oxide will react You only need to calculate based on copper oxide You don't need to find the mass of carbon Common Question Types Type 1: "What mass of product is formed from X g of reactant?" Strategy: Reactant mass → reactant moles → product moles → product mass Type 2: "What mass of reactant is needed to make X g of product?" Strategy: Product mass → product moles → reactant moles → reactant mass Type 3: "What mass of B reacts with X g of A?" Strategy: A mass → A moles → B moles → B mass Conservation of Mass Remember that mass is conserved in chemical reactions: Total mass of reactants = Total mass of products This can be a useful check for your answers!
Example: If 12 g Mg reacts with 8 g O₂ to make 20 g MgO Reactants: 12 + 8 = 20 g…
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