Limiting Reagents
Worksheet
Easy Limited Reagent Reacting Mass Questions
1. 5g of magnesium are oxidised in 10 grams of oxygen. What mass of magnesium oxide will it produce?
2Mg + O2
→
2MgO
Working Out
Moles Mg = 5 / 24.3 = 0.206 mol
Moles O2 = 10 / 32.0 = 0.313 mol
Ratio Mg : O2 = 2 : 1.
0.206 mol Mg needs 0.103 mol O2.
We have 0.313 mol (Excess).
Limiting Reagent: Magnesium
Mass MgO = 0.206 × 40.3 = 8.30 g
(Question value ‘5g’ has 1 sig fig)
Answer = 8 g
Moles O2 = 10 / 32.0 = 0.313 mol
Ratio Mg : O2 = 2 : 1.
0.206 mol Mg needs 0.103 mol O2.
We have 0.313 mol (Excess).
Limiting Reagent: Magnesium
Mass MgO = 0.206 × 40.3 = 8.30 g
(Question value ‘5g’ has 1 sig fig)
Answer = 8 g
2. 12g of methane are combusted in 14g of oxygen. State the mass of carbon dioxide produced in the reaction. Assume that complete combustion takes place.
CH4 + 2O2
→
CO2 + 2H2O
Working Out
Moles CH4 = 12 / 16.0 = 0.75 mol
Moles O2 = 14 / 32.0 = 0.438 mol
Ratio 1 : 2. 0.75 mol CH4 needs 1.50 mol O2.
We have 0.438 mol. O2 is Limiting.
Moles CO2 = 0.438 / 2 = 0.219 mol.
Mass CO2 = 0.219 × 44.0 = 9.64 g
(12g and 14g have 2 s.f.)
Answer = 9.6 g
Moles O2 = 14 / 32.0 = 0.438 mol
Ratio 1 : 2. 0.75 mol CH4 needs 1.50 mol O2.
We have 0.438 mol. O2 is Limiting.
Moles CO2 = 0.438 / 2 = 0.219 mol.
Mass CO2 = 0.219 × 44.0 = 9.64 g
(12g and 14g have 2 s.f.)
Answer = 9.6 g
3. 36g of methane and 32g of water are reacted together. State the mass of hydrogen produced in the reaction.
CH4 + H2O
→
CO + 3H2
Working Out
Moles CH4 = 36 / 16.0 = 2.25 mol
Moles H2O = 32 / 18.0 = 1.78 mol
Ratio 1 : 1. H2O is Limiting.
Ratio H2O : H2 = 1 : 3.
Moles H2 = 1.78 × 3 = 5.34 mol.
Mass H2 = 5.34 × 2.0 = 10.7 g
(36g and 32g have 2 s.f.)
Answer = 11 g
Moles H2O = 32 / 18.0 = 1.78 mol
Ratio 1 : 1. H2O is Limiting.
Ratio H2O : H2 = 1 : 3.
Moles H2 = 1.78 × 3 = 5.34 mol.
Mass H2 = 5.34 × 2.0 = 10.7 g
(36g and 32g have 2 s.f.)
Answer = 11 g
4. 60g of propene and 42g of chlorine react. State the mass of 1,2-dichloropropane formed.
C3H6 + Cl2
→
C3H6Cl2
Working Out
Moles C3H6 = 60 / 42.0 = 1.43 mol
Moles Cl2 = 42 / 71.0 = 0.592 mol
Ratio 1 : 1. Cl2 is Limiting.
Moles Product = 0.592 mol.
Mr C3H6Cl2 = 113.0
Mass = 0.592 × 113.0 = 66.9 g
(60g and 42g have 2 s.f.)
Answer = 67 g
Moles Cl2 = 42 / 71.0 = 0.592 mol
Ratio 1 : 1. Cl2 is Limiting.
Moles Product = 0.592 mol.
Mr C3H6Cl2 = 113.0
Mass = 0.592 × 113.0 = 66.9 g
(60g and 42g have 2 s.f.)
Answer = 67 g
5. 100g of potassium reacts with 80g of iodine. State the mass of potassium iodide formed.
2K + I2
→
2KI
Working Out
Moles K = 100 / 39.1 = 2.56 mol
Moles I2 = 80 / 253.8 = 0.315 mol
Ratio 2 : 1. 0.315 mol I2 needs 0.63 mol K.
We have 2.56 mol (Excess). I2 is Limiting.
Moles KI = 0.315 × 2 = 0.630 mol.
Mass KI = 0.630 × 166.0 = 104.6 g
(100g and 80g have 2 s.f.)
Answer = 100 g (1.0 × 10² g)
Moles I2 = 80 / 253.8 = 0.315 mol
Ratio 2 : 1. 0.315 mol I2 needs 0.63 mol K.
We have 2.56 mol (Excess). I2 is Limiting.
Moles KI = 0.315 × 2 = 0.630 mol.
Mass KI = 0.630 × 166.0 = 104.6 g
(100g and 80g have 2 s.f.)
Answer = 100 g (1.0 × 10² g)
Medium Difficulty Limited Reagent Reacting Mass Questions
1. A heating unit has 60l of liquified propane gas and a separate oxygen supply that contains 1000l of oxygen stored at 10 atmospheres. If LPG has a density of 0.5kg/l, what mass of carbon dioxide will it produce?
Working Out
C3H8 + 5O2
→
3CO2 + 4H2O
Propane (LPG): Mass = 60 L × 0.5 kg/L = 30 kg = 30,000 g.
Moles = 30,000 / 44.1 = 680.3 mol.
Oxygen: pV = nRT.
p = 10 atm = 1,013,250 Pa. V = 1000 L = 1.0 m³. T = 298 K.
n = (1,013,250 × 1.0) / (8.31 × 298) = 409.1 mol.
Limiting: Ratio 1 : 5. 680 mol C3H8 needs 3400 mol O2.
We have 409.1 mol. Oxygen is Limiting.
Ratio O2 : CO2 = 5 : 3.
Moles CO2 = 409.1 × (3/5) = 245.5 mol.
Mass CO2 = 245.5 × 44.0 = 10,802 g = 10.8 kg.
(60l has 2 s.f.)
Answer = 11 kg
Moles = 30,000 / 44.1 = 680.3 mol.
Oxygen: pV = nRT.
p = 10 atm = 1,013,250 Pa. V = 1000 L = 1.0 m³. T = 298 K.
n = (1,013,250 × 1.0) / (8.31 × 298) = 409.1 mol.
Limiting: Ratio 1 : 5. 680 mol C3H8 needs 3400 mol O2.
We have 409.1 mol. Oxygen is Limiting.
Ratio O2 : CO2 = 5 : 3.
Moles CO2 = 409.1 × (3/5) = 245.5 mol.
Mass CO2 = 245.5 × 44.0 = 10,802 g = 10.8 kg.
(60l has 2 s.f.)
Answer = 11 kg
2. A combustion engine releases 20dm³ of nitrogen monoxide and 15dm³ of carbon monoxide. Both are passed over a catalytic converter, where they react to produce nitrogen and carbon dioxide. Which of the two pollutants will the car release into the environment, and how much will it release?
2NO + 2CO
→
N2 + 2CO2
Working Out
Avogadro’s Law: Volume ∝ Moles.
Ratio NO : CO = 2 : 2 (or 1 : 1).
15 dm³ CO reacts with 15 dm³ NO.
We have 20 dm³ NO.
CO is Limiting. NO is in Excess.
Excess NO = 20 – 15 = 5 dm³.
(20 and 15 have 2 s.f., result 5 has 1 s.f.)
Answer = 5 dm³ of Nitrogen Monoxide (NO)
Ratio NO : CO = 2 : 2 (or 1 : 1).
15 dm³ CO reacts with 15 dm³ NO.
We have 20 dm³ NO.
CO is Limiting. NO is in Excess.
Excess NO = 20 – 15 = 5 dm³.
(20 and 15 have 2 s.f., result 5 has 1 s.f.)
Answer = 5 dm³ of Nitrogen Monoxide (NO)
3. 1000 tonnes of iron oxide are reduced using carbon monoxide. 240 000 m³ of carbon monoxide are used. State the mass of iron produced.
Working Out
Fe2O3 + 3CO
→
2Fe + 3CO2
Iron Oxide: Mass = 1000 × 106 g.
Moles = 109 / 159.6 = 6.27 × 106 mol.
Carbon Monoxide: pV = nRT (p=100kPa, T=298K).
n = (100,000 × 240,000) / (8.31 × 298) = 9.69 × 106 mol.
Limiting: Ratio 1 : 3. Fe2O3 needs 18.8 × 106 mol CO.
We have 9.69 × 106 mol. CO is Limiting.
Ratio CO : Fe = 3 : 2.
Moles Fe = 9.69 × 106 × (2/3) = 6.46 × 106 mol.
Mass Fe = 6.46 × 106 × 55.8 = 360.5 tonnes.
(240,000 implies 2 s.f.)
Answer = 370 tonnes
Moles = 109 / 159.6 = 6.27 × 106 mol.
Carbon Monoxide: pV = nRT (p=100kPa, T=298K).
n = (100,000 × 240,000) / (8.31 × 298) = 9.69 × 106 mol.
Limiting: Ratio 1 : 3. Fe2O3 needs 18.8 × 106 mol CO.
We have 9.69 × 106 mol. CO is Limiting.
Ratio CO : Fe = 3 : 2.
Moles Fe = 9.69 × 106 × (2/3) = 6.46 × 106 mol.
Mass Fe = 6.46 × 106 × 55.8 = 360.5 tonnes.
(240,000 implies 2 s.f.)
Answer = 370 tonnes
4. Sulfur hexafluoride is synthesised by the reaction between cobalt (III) fluoride and sulfur tetrafluoride. In this reaction, cobalt (III) fluoride is reduced to cobalt (II) chloride and sulfur tetrafluoride is oxidised to sulfur hexafluoride. If you start with 12 grams of cobalt (III) fluoride and 16 grams of sulfur tetrafluoride, what mass of sulfur hexafluoride will you make?
Working Out
2CoF3 + SF4
→
2CoF2 + SF6
Moles CoF3 = 12 / 115.9 = 0.104 mol.
Moles SF4 = 16 / 108.1 = 0.148 mol.
Ratio 2 : 1. 0.104 mol CoF3 needs 0.052 mol SF4.
We have 0.148 mol (Excess). CoF3 is Limiting.
Ratio CoF3 : SF6 = 2 : 1.
Moles SF6 = 0.104 / 2 = 0.052 mol.
Mass SF6 = 0.052 × 146.1 = 7.60 g.
(12g has 2 s.f.)
Answer = 7.6 g
Moles SF4 = 16 / 108.1 = 0.148 mol.
Ratio 2 : 1. 0.104 mol CoF3 needs 0.052 mol SF4.
We have 0.148 mol (Excess). CoF3 is Limiting.
Ratio CoF3 : SF6 = 2 : 1.
Moles SF6 = 0.104 / 2 = 0.052 mol.
Mass SF6 = 0.052 × 146.1 = 7.60 g.
(12g has 2 s.f.)
Answer = 7.6 g
Hard Limiting Reagent Reacting Mass Questions
1. A coal power station releases sulfur dioxide in the exhaust gases due to sulfur impurities in the fuel. The sulfur dioxide is removed through neutralisation. The gas is bubbled through water to produce an acid solution, which is reacted against calcium hydroxide. The power station oxidises 30 tonnes of sulfur per day and has 30 tonnes of calcium hydroxide added to the flue gas per day. State the volume of sulfur dioxide that is still released in the atmosphere.
Working Out
Step 1: S → SO2. 30t S → 30t SO2 (moles).
Moles S = 30 × 106 / 32.1 = 934,579 mol = Moles SO2 Produced.
Step 2: SO2 + Ca(OH)2 → CaSO3 + H2O.
Moles Ca(OH)2 = 30 × 106 / 74.1 = 404,858 mol.
Ratio 1 : 1. Ca(OH)2 is Limiting.
Moles SO2 Removed = 404,858 mol.
Step 3: Remaining SO2.
934,579 – 404,858 = 529,721 mol.
Volume = nRT / p = (529,721 × 8.31 × 298) / 100,000 = 13,118 m³.
(30 tonnes has 2 s.f.)
Answer = 1.3 × 107 dm³ (or 13,000 m³)
Moles S = 30 × 106 / 32.1 = 934,579 mol = Moles SO2 Produced.
Step 2: SO2 + Ca(OH)2 → CaSO3 + H2O.
Moles Ca(OH)2 = 30 × 106 / 74.1 = 404,858 mol.
Ratio 1 : 1. Ca(OH)2 is Limiting.
Moles SO2 Removed = 404,858 mol.
Step 3: Remaining SO2.
934,579 – 404,858 = 529,721 mol.
Volume = nRT / p = (529,721 × 8.31 × 298) / 100,000 = 13,118 m³.
(30 tonnes has 2 s.f.)
Answer = 1.3 × 107 dm³ (or 13,000 m³)
2. A hydrocarbon with the molar mass of 68 was reacted against 2dm³ of chlorine gas, producing 2.78g of product. What are the possible skeletal formulae of the hydrocarbon?
Working Out
Mr 68 = C5H8 (Formula CnH2n-2).
Moles Chlorine: pV=nRT.
V = 0.002 m³. n = (100,000 × 0.002) / (8.31 × 298) = 0.0808 mol.
Moles Product: Assuming addition (C5H8Cl2, Mr=139).
n = 2.78 / 139 = 0.020 mol.
Since 0.020 < 0.0808, Chlorine was in excess, Hydrocarbon was limiting.
Formula fits C5H8 (Alkyne or Diene).
Answer: Isomers of C5H8
Moles Chlorine: pV=nRT.
V = 0.002 m³. n = (100,000 × 0.002) / (8.31 × 298) = 0.0808 mol.
Moles Product: Assuming addition (C5H8Cl2, Mr=139).
n = 2.78 / 139 = 0.020 mol.
Since 0.020 < 0.0808, Chlorine was in excess, Hydrocarbon was limiting.
Formula fits C5H8 (Alkyne or Diene).
Answer: Isomers of C5H8
- Pent-1-yne
- Pent-2-yne
- 3-methylbut-1-yne
- Penta-1,3-diene
- Penta-1,4-diene
- Isoprene (2-methylbuta-1,3-diene)
- Cyclopentene