Reacting Masses
Worksheet
Easy Questions
1. 5g of magnesium are oxidised in excess oxygen. What mass of magnesium oxide will it produce?
2Mg + O2
→
2MgO
Working Out
Moles Mg = 5 / 24.3 = 0.206 mol
Ratio Mg : MgO = 1 : 1
Mass MgO = 0.206 × (24.3 + 16.0)
Mass MgO = 0.206 × 40.3 = 8.29 g
(Question value ‘5g’ has 1 sig fig)
Answer = 8 g
Ratio Mg : MgO = 1 : 1
Mass MgO = 0.206 × (24.3 + 16.0)
Mass MgO = 0.206 × 40.3 = 8.29 g
(Question value ‘5g’ has 1 sig fig)
Answer = 8 g
2. 36g of copper carbonate are thermally decomposed to form copper oxide.
CuCO3
→
CuO + CO2
a) What mass of copper oxide will it produce?
Moles CuCO3 = 36 / (63.5 + 12 + 48) = 0.291 mol
Ratio CuCO3 : CuO = 1 : 1
Mass CuO = 0.291 × (63.5 + 16.0) = 0.291 × 79.5 = 23.15 g
(Question value ’36g’ has 2 sig figs)
Answer = 23 g
Ratio CuCO3 : CuO = 1 : 1
Mass CuO = 0.291 × (63.5 + 16.0) = 0.291 × 79.5 = 23.15 g
(Question value ’36g’ has 2 sig figs)
Answer = 23 g
b) What mass of carbon dioxide will it produce?
Ratio CuCO3 : CO2 = 1 : 1
Mass CO2 = 0.291 × (12.0 + 32.0) = 0.291 × 44.0 = 12.81 g
(Question value ’36g’ has 2 sig figs)
Answer = 13 g
Mass CO2 = 0.291 × (12.0 + 32.0) = 0.291 × 44.0 = 12.81 g
(Question value ’36g’ has 2 sig figs)
Answer = 13 g
3. 62g of ammonia are reacted with excess sulfuric acid. State the mass of ammonium sulfate produced.
2NH3 + H2SO4
→
(NH4)2SO4
Working Out
Moles NH3 = 62 / 17 = 3.65 mol
Ratio NH3 : Salt = 2 : 1
Moles Salt = 3.65 / 2 = 1.82 mol
Mass = 1.82 × 132.1 = 240.6 g
(Question value ’62g’ has 2 sig figs)
Answer = 240 g
Ratio NH3 : Salt = 2 : 1
Moles Salt = 3.65 / 2 = 1.82 mol
Mass = 1.82 × 132.1 = 240.6 g
(Question value ’62g’ has 2 sig figs)
Answer = 240 g
4. 134g of methane is fully combusted. State the mass of carbon dioxide formed.
CH4 + 2O2
→
CO2 + 2H2O
Working Out
Moles CH4 = 134 / 16 = 8.38 mol
Ratio 1 : 1
Mass CO2 = 8.38 × 44.0 = 368.5 g
(Question value ‘134g’ has 3 sig figs)
Answer = 369 g
Ratio 1 : 1
Mass CO2 = 8.38 × 44.0 = 368.5 g
(Question value ‘134g’ has 3 sig figs)
Answer = 369 g
5. State the mass of nitrogen required to form 120g of ammonia using the Haber process.
N2 + 3H2
→
2NH3
Working Out
Moles NH3 = 120 / 17 = 7.06 mol
Ratio N2 : NH3 = 1 : 2
Moles N2 = 7.06 / 2 = 3.53 mol
Mass N2 = 3.53 × 28.0 = 98.8 g
(Question value ‘120g’ has 3 sig figs as per new rule)
Answer = 98.8 g
Ratio N2 : NH3 = 1 : 2
Moles N2 = 7.06 / 2 = 3.53 mol
Mass N2 = 3.53 × 28.0 = 98.8 g
(Question value ‘120g’ has 3 sig figs as per new rule)
Answer = 98.8 g
6. 7 moles of barium nitrate are added to excess sodium sulfate solution. State the mass of precipitate formed.
Ba(NO3)2 + Na2SO4
→
2NaNO3 + BaSO4
Working Out
Ratio 1 : 1 so Moles Precipitate (BaSO4) = 7 mol
Mr BaSO4 = 137.3 + 32.1 + 64 = 233.4
Mass = 7 × 233.4 = 1634 g
(Question value ‘7 moles’ has 1 sig fig)
Answer = 2000 g (or 2 kg)
Mr BaSO4 = 137.3 + 32.1 + 64 = 233.4
Mass = 7 × 233.4 = 1634 g
(Question value ‘7 moles’ has 1 sig fig)
Answer = 2000 g (or 2 kg)
7. What mass of magnesium do you need to use to produce 60 tonnes of titanium?
2Mg + TiCl4
→
2MgCl2 + Ti
Working Out
Moles Ti = 60,000,000 / 47.9 = 1,252,609 mol
Ratio Mg : Ti = 2 : 1
Moles Mg = 2,505,219 mol
Mass Mg = 2,505,219 × 24.3 = 60,876,826 g = 60.9 tonnes
(Question value ’60 tonnes’ has 2 sig figs as per new rule)
Answer = 61 tonnes
Ratio Mg : Ti = 2 : 1
Moles Mg = 2,505,219 mol
Mass Mg = 2,505,219 × 24.3 = 60,876,826 g = 60.9 tonnes
(Question value ’60 tonnes’ has 2 sig figs as per new rule)
Answer = 61 tonnes
8. What mass of oxygen would the decomposition of 12 grams of hydrogen peroxide produce?
2H2O2
→
2H2O + O2
Working Out
Moles H2O2 = 12 / 34 = 0.353 mol
Ratio H2O2 : O2 = 2 : 1
Moles O2 = 0.353 / 2 = 0.177 mol
Mass = 0.177 × 32.0 = 5.65 g
(Question value ’12g’ has 2 sig figs)
Answer = 5.7 g
Ratio H2O2 : O2 = 2 : 1
Moles O2 = 0.353 / 2 = 0.177 mol
Mass = 0.177 × 32.0 = 5.65 g
(Question value ’12g’ has 2 sig figs)
Answer = 5.7 g
Medium Questions
1. A heating unit combusts 100 litres of liquified propane (density 0.5kg/l). What mass of carbon dioxide is produced?
Working Out
C3H8 + 5O2
→
3CO2 + 4H2O
Mass propane = 0.5 × 100 = 50 kg = 50,000 g
Moles Propane = 50,000 / 44.1 = 1134 mol
Ratio 1 : 3 → Moles CO2 = 3401 mol
Mass CO2 = 3401 × 44.0 = 149,644 g
(Question values ‘100L’ has 3 sig figs)
Answer = 150 kg
Moles Propane = 50,000 / 44.1 = 1134 mol
Ratio 1 : 3 → Moles CO2 = 3401 mol
Mass CO2 = 3401 × 44.0 = 149,644 g
(Question values ‘100L’ has 3 sig figs)
Answer = 150 kg
2. 40 tonnes of aluminium oxide are electrolysed.
2Al2O3
→
4Al + 3O2
(Moles Al2O3 = 40×106 / 102 = 392,157 mol)
a) What mass of aluminium will it produce?
Ratio 2:4 (1:2)
Moles Al = 392,157 × 2 = 784,314 mol
Mass = 784,314 × 27.0 = 21.2 tonnes
(Question value ’40 tonnes’ has 2 sig figs)
Answer = 21 tonnes
Moles Al = 392,157 × 2 = 784,314 mol
Mass = 784,314 × 27.0 = 21.2 tonnes
(Question value ’40 tonnes’ has 2 sig figs)
Answer = 21 tonnes
b) What mass of carbon dioxide will it produce?
Reaction at anode: C + O2 → CO2
Ratio Al2O3 : O2 is 2:3
Moles O2 = 392,157 × 1.5 = 588,235 mol
Mass CO2 = 588,235 × 44.0 = 25.9 tonnes
(Question value ’40 tonnes’ has 2 sig figs)
Answer = 26 tonnes
Ratio Al2O3 : O2 is 2:3
Moles O2 = 392,157 × 1.5 = 588,235 mol
Mass CO2 = 588,235 × 44.0 = 25.9 tonnes
(Question value ’40 tonnes’ has 2 sig figs)
Answer = 26 tonnes
3. 90 tonnes of iron are extracted from iron (III) oxide by reduction using carbon monoxide. What mass of iron (III) oxide did you start with?
Working Out
Fe2O3 + 3CO
→
2Fe + 3CO2
Moles Fe = 90,000,000 / 55.8 = 1,612,903 mol
Ratio Fe : Fe2O3 = 2 : 1
Moles Oxide = 806,452 mol
Mass = 806,452 × 159.6 = 128.7 tonnes
(Question value ’90 tonnes’ has 2 sig figs)
Answer = 130 tonnes
Ratio Fe : Fe2O3 = 2 : 1
Moles Oxide = 806,452 mol
Mass = 806,452 × 159.6 = 128.7 tonnes
(Question value ’90 tonnes’ has 2 sig figs)
Answer = 130 tonnes
4. How many moles of sulfuric acid will you make if you start with 20kg of sulfur?
Working Out
The reaction sequence S → SO2 → SO3 → H2SO4 maintains a 1:1 atomic ratio of Sulfur.
Moles S = 20,000 / 32.1 = 623 mol
(Question value ’20kg’ has 2 sig figs)
Answer = 620 moles
(Question value ’20kg’ has 2 sig figs)
Answer = 620 moles
5. Sulfur hexafluoride synthesis from 12g of cobalt (III) fluoride. What mass of SF6 is made?
Working Out
2CoF3 + SF4
→
2CoF2 + SF6
Moles CoF3 = 12 / 115.9 = 0.104 mol
Ratio CoF3 : SF6 = 2 : 1
Moles SF6 = 0.052 mol
Mass = 0.052 × 146.1 = 7.56 g
(Question value ’12g’ has 2 sig figs)
Answer = 7.6 g
Ratio CoF3 : SF6 = 2 : 1
Moles SF6 = 0.052 mol
Mass = 0.052 × 146.1 = 7.56 g
(Question value ’12g’ has 2 sig figs)
Answer = 7.6 g
Hard Questions
1. 60L of LPG (propane) combusts producing 45kg of CO2. What mass of CO is produced?
Working Out
Mass Propane = 60 × 0.5 = 30 kg = 30,000 g
Moles Propane = 30,000 / 44.1 = 680.3 mol
Total Moles of C atoms = 680.3 × 3 = 2041 mol
Moles CO2 produced = 45,000 / 44.0 = 1023 mol
(This accounts for 1023 moles of C)
Remaining C atoms = 2041 – 1023 = 1018 mol
Since no soot, remaining C is CO.
Mass CO = 1018 × 28.0 = 28,504 g = 28.5 kg
(Question values 60 & 45 have 2 sig figs)
Answer = 29 kg
Moles Propane = 30,000 / 44.1 = 680.3 mol
Total Moles of C atoms = 680.3 × 3 = 2041 mol
Moles CO2 produced = 45,000 / 44.0 = 1023 mol
(This accounts for 1023 moles of C)
Remaining C atoms = 2041 – 1023 = 1018 mol
Since no soot, remaining C is CO.
Mass CO = 1018 × 28.0 = 28,504 g = 28.5 kg
(Question values 60 & 45 have 2 sig figs)
Answer = 29 kg
2. State the mass of calcium hydroxide required to neutralise 400 tonnes of sulfur impurities.
Working Out
Assuming SO2 + Ca(OH)2 → CaSO3 + H2O (Ratio 1:1)
Moles S = 400×106 / 32.1 = 12.46×106 mol
Moles Ca(OH)2 = 12.46×106 mol
Mass = 12.46×106 × 74.1 = 923 tonnes
(Note: If calculating based on SO2 mass directly: 400t SO2 -> 462t Ca(OH)2. Assuming ‘sulfur impurities’ refers to mass of elemental S: 923t)
Standard Answer (using SO2 mass): 462 tonnes
(Question value ‘400 tonnes’ has 3 sig figs)
Answer based on S impurity = 923 tonnes
Moles S = 400×106 / 32.1 = 12.46×106 mol
Moles Ca(OH)2 = 12.46×106 mol
Mass = 12.46×106 × 74.1 = 923 tonnes
(Note: If calculating based on SO2 mass directly: 400t SO2 -> 462t Ca(OH)2. Assuming ‘sulfur impurities’ refers to mass of elemental S: 923t)
Standard Answer (using SO2 mass): 462 tonnes
(Question value ‘400 tonnes’ has 3 sig figs)
Answer based on S impurity = 923 tonnes
3. What is the net change in carbon dioxide if you produce 10kg of ethene from fermentation?
Working Out
Fermentation produces 2 moles CO2 for every 2 moles Ethanol. Dehydration produces 1 mole Ethene per 1 mole Ethanol. Thus, 1 mole Ethene produced = 1 mole CO2 released.
Moles Ethene (C2H4) = 10,000 / 28.0 = 357 mol
Moles CO2 = 357 mol
Mass CO2 = 357 × 44.0 = 15,708 g = 15.7 kg
(Question value ’10kg’ has 2 sig figs)
Answer = 16 kg
Moles CO2 = 357 mol
Mass CO2 = 357 × 44.0 = 15,708 g = 15.7 kg
(Question value ’10kg’ has 2 sig figs)
Answer = 16 kg
4. 5g tablet (MgCO3 + Ca(OH)2). Used 0.06 mol HCl, produced 0.4g CO2. State masses in tablet.
Working Out
Step 1: MgCO3
Moles CO2 = 0.4 / 44.0 = 0.0091 mol
Moles MgCO3 = 0.0091 mol
Mass MgCO3 = 0.0091 × 84.3 = 0.77 g
(Inputs have 1 sig fig)
Answer = 0.8 g
Step 2: Ca(OH)2
HCl used by MgCO3 (Ratio 1:2) = 0.0182 mol.
Remaining HCl = 0.06 – 0.0182 = 0.0418 mol.
Moles Ca(OH)2 (Ratio 1:2) = 0.0209 mol.
Mass Ca(OH)2 = 0.0209 × 74.1 = 1.55 g
Answer = 2 g
Moles CO2 = 0.4 / 44.0 = 0.0091 mol
Moles MgCO3 = 0.0091 mol
Mass MgCO3 = 0.0091 × 84.3 = 0.77 g
(Inputs have 1 sig fig)
Answer = 0.8 g
Step 2: Ca(OH)2
HCl used by MgCO3 (Ratio 1:2) = 0.0182 mol.
Remaining HCl = 0.06 – 0.0182 = 0.0418 mol.
Moles Ca(OH)2 (Ratio 1:2) = 0.0209 mol.
Mass Ca(OH)2 = 0.0209 × 74.1 = 1.55 g
Answer = 2 g
5. 40g alkane reacted with excess chlorine. HCl titrated. Calculate the formula.
Working Out
Moles NaOH used = 4 × (17.24/1000) = 0.06896 mol
So, 0.06896 mol HCl in 10cm3 sample.
Total HCl in 1000cm3 = 6.896 mol.
Testing Alkanes (40g):
If Butane (C4H10, Mr 58): 40/58 = 0.69 mol.
Ratio HCl / Alkane = 6.896 / 0.69 = 10.
Reaction: C4H10 + 10Cl2 → C4Cl10 + 10HCl
This fits perfectly (replacing all 10 Hydrogens).
Answer = Butane (C4H10)
So, 0.06896 mol HCl in 10cm3 sample.
Total HCl in 1000cm3 = 6.896 mol.
Testing Alkanes (40g):
If Butane (C4H10, Mr 58): 40/58 = 0.69 mol.
Ratio HCl / Alkane = 6.896 / 0.69 = 10.
Reaction: C4H10 + 10Cl2 → C4Cl10 + 10HCl
This fits perfectly (replacing all 10 Hydrogens).
Answer = Butane (C4H10)