Unit 3: Chemical Equations and Reactions

Using balanced equations to predict and calculate the outcomes of chemical changes.

3.20 Introducing Chemical Equations

A chemical equation is a shorthand representation of a chemical reaction. It shows the substances that react, called reactants, and the new substances that are formed, called products. The arrow ($\rightarrow$) indicates the direction of the reaction.

  • Reactants: The starting substances in a reaction, written on the left side of the equation.
  • Products: The new substances created in a reaction, written on the right side of the equation.
  • Stoichiometric Coefficients: The numbers written in front of chemical formulae that show the mole ratio of reactants and products.
  • State Symbols: Indicate the physical state of each substance: (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous (dissolved in water).

Consider the equation for photosynthesis: $6CO_2(g) + 6H_2O(l) \rightarrow C_6H_{12}O_6(aq) + 6O_2(g)$. This tells us that 6 moles of carbon dioxide gas react with 6 moles of liquid water to produce 1 mole of aqueous glucose and 6 moles of oxygen gas.

Solved Examples:
  1. In the reaction $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$, identify the reactants and products.
    Solution: The reactants are hydrogen ($H_2$) and oxygen ($O_2$). The product is water ($H_2O$).
  2. What is the mole ratio of magnesium to hydrogen in the equation $Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)$?
    Solution: The stoichiometric coefficient for Mg is 1 (unwritten) and for Hâ‚‚ is 1. The mole ratio is 1:1.
  3. For the reaction above, how many moles of HCl are needed to react with 1 mole of Mg?
    Solution: The mole ratio of Mg to HCl is 1:2. Therefore, 2 moles of HCl are needed.
  4. What do the state symbols in $AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)$ mean?
    Solution: (aq) means the substance is dissolved in water (aqueous), and (s) means the substance is a solid (a precipitate in this case).
  5. In the Haber process, $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$, how many moles of ammonia ($NH_3$) are produced from 1 mole of nitrogen ($N_2$)?
    Solution: The mole ratio of N₂ to NH₃ is 1:2. Therefore, 2 moles of ammonia are produced.
  6. Identify the stoichiometric coefficients in the equation $4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(s)$.
    Solution: The stoichiometric coefficients are 4, 3, and 2.
  7. If 0.5 moles of $H_2S$ react in $2H_2S + 3O_2 \rightarrow 2SO_2 + 2H_2O$, how many moles of $O_2$ are required?
    Solution: The ratio of $H_2S$ to $O_2$ is 2:3. Moles of $O_2 = 0.5 \times \frac{3}{2} = 0.75$ moles.
  8. What does the formula $C_6H_{12}O_6(aq)$ tell you about the substance?
    Solution: It is a molecule containing 6 carbon, 12 hydrogen, and 6 oxygen atoms, and it is dissolved in water.
  9. From the equation $2K + 2H_2O \rightarrow 2KOH + H_2$, what is the ratio of moles of potassium (K) used to moles of hydrogen ($H_2$) produced?
    Solution: The ratio is 2 moles of K to 1 mole of Hâ‚‚, or 2:1.
  10. If a reaction produces 4 moles of NOâ‚‚ in $2Pb(NO_3)_2(s) \rightarrow 2PbO(s) + 4NO_2(g) + O_2(g)$, how many moles of Oâ‚‚ are also produced?
    Solution: The ratio of NOâ‚‚ to Oâ‚‚ is 4:1. Therefore, 1 mole of Oâ‚‚ is produced.

3.21 The Law of Conservation of Mass & Balancing Equations

The Law of Conservation of Mass states that in a closed system, the total mass of the reactants must equal the total mass of the products. This is because atoms are merely rearranged during a chemical reaction, not created or destroyed. To uphold this law, chemical equations must be balanced, meaning there must be an equal number of each type of atom on both sides of the equation. This is achieved by adjusting the stoichiometric coefficients.

A simple strategy for balancing equations:
  1. Write the unbalanced equation with the correct chemical formulae.
  2. Balance elements that appear in only one compound on each side first.
  3. Balance polyatomic ions (like $SO_4^{2-}$) as a single unit if they appear on both sides.
  4. Balance elements that appear in their pure form (e.g., $O_2$, $Na$) last.
  5. Check that the number of atoms of each element is the same on both sides.
Solved Examples:
  1. Balance the equation: $N_2 + H_2 \rightarrow NH_3$.
    Solution: Balance N: $N_2 + H_2 \rightarrow 2NH_3$. Now there are 6 H on the right, so balance H: $N_2 + 3H_2 \rightarrow 2NH_3$. The equation is balanced.
  2. Balance the equation: $CH_4 + O_2 \rightarrow CO_2 + H_2O$.
    Solution: C is balanced. Balance H: $CH_4 + O_2 \rightarrow CO_2 + 2H_2O$. Now there are 4 O on the right, so balance O: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$. The equation is balanced.
  3. Balance the equation: $Al + Cl_2 \rightarrow AlCl_3$.
    Solution: Balance Cl using a common multiple (6): $Al + 3Cl_2 \rightarrow 2AlCl_3$. Now balance Al: $2Al + 3Cl_2 \rightarrow 2AlCl_3$. The equation is balanced.
  4. Balance the equation: $Fe_2O_3 + C \rightarrow Fe + CO_2$.
    Solution: Balance Fe: $Fe_2O_3 + C \rightarrow 2Fe + CO_2$. Balance O using multiples: $2Fe_2O_3 + C \rightarrow 4Fe + 3CO_2$. Now balance C: $2Fe_2O_3 + 3C \rightarrow 4Fe + 3CO_2$. It is balanced.
  5. Balance the equation: $Na + O_2 \rightarrow Na_2O$.
    Solution: Balance O: $Na + O_2 \rightarrow 2Na_2O$. Now balance Na: $4Na + O_2 \rightarrow 2Na_2O$. It is balanced.
  6. Balance the equation for the combustion of propane: $C_3H_8 + O_2 \rightarrow CO_2 + H_2O$.
    Solution: Balance C: $C_3H_8 + O_2 \rightarrow 3CO_2 + H_2O$. Balance H: $C_3H_8 + O_2 \rightarrow 3CO_2 + 4H_2O$. Balance O: Right side has $6+4=10$ O atoms. $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$.
  7. Balance: $H_2SO_4 + NaOH \rightarrow Na_2SO_4 + H_2O$.
    Solution: Balance Na: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + H_2O$. Now balance H: Left has 2+2=4 H. Right has 2 H. $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$. It is balanced.
  8. Balance: $Ca(OH)_2 + H_3PO_4 \rightarrow Ca_3(PO_4)_2 + H_2O$.
    Solution: Balance Ca and POâ‚„ groups first: $3Ca(OH)_2 + 2H_3PO_4 \rightarrow Ca_3(PO_4)_2 + H_2O$. Now count H atoms: Left has $3 \times 2 + 2 \times 3 = 12$ H. Balance H: $3Ca(OH)_2 + 2H_3PO_4 \rightarrow Ca_3(PO_4)_2 + 6H_2O$.
  9. Balance: $Cu + HNO_3 \rightarrow Cu(NO_3)_2 + NO_2 + H_2O$.
    Solution: This is a redox reaction and is trickier. Balance Cu (it's 1:1). Trial and error leads to: $Cu + 4HNO_3 \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O$.
  10. Balance: $KClO_3 \rightarrow KCl + O_2$.
    Solution: The oxygen is unbalanced (3 on left, 2 on right). Find a common multiple (6). $2KClO_3 \rightarrow KCl + 3O_2$. Now balance K and Cl: $2KClO_3 \rightarrow 2KCl + 3O_2$. It is balanced.

3.22 Calculating Reacting Quantities (Stoichiometry)

Stoichiometry is the area of chemistry that uses the relationships in a balanced chemical equation to calculate the quantities of reactants and products. A balanced equation acts as a chemical "recipe," and the mole ratio derived from its coefficients is the key conversion factor that links every substance in the reaction.

No matter what quantity you are given (mass, gas volume, solution volume), the strategy is always to convert it into moles first. The mole is the central unit that connects all these different measurements.

The Core Stoichiometry Strategy:
  1. Balanced Equation: Ensure you have a correctly balanced chemical equation. This is non-negotiable as it provides the correct mole ratios.
  2. Moles of Given: Convert the quantity of the substance you are given (the "known") into moles.
    • From mass: $n = \frac{\text{mass}}{M_r}$
    • From solution: $n = C \times V$ (V in dm³)
    • From gas: $n = \frac{PV}{RT}$ (V in m³)
  3. Mole Ratio Bridge: Use the stoichiometric coefficients from the balanced equation to create a mole ratio. This ratio acts as a bridge to find the moles of the substance you want to find (the "unknown"). $$ \text{moles of unknown} = \text{moles of known} \times \frac{\text{coefficient of unknown}}{\text{coefficient of known}} $$
  4. Calculate Required Quantity: Convert the moles of the unknown substance back into the desired quantity (mass, volume, concentration, etc.).
Solved Examples:
  1. What mass of potassium oxide ($K_2O$) is formed when 7.8 g of potassium is burned in excess oxygen? $4K + O_2 \rightarrow 2K_2O$.
    Solution: Moles K = $\frac{7.8}{39.1} \approx 0.2$ mol. Mole ratio K:$K_2O$ is 4:2. Moles $K_2O = 0.2 \times \frac{2}{4} = 0.1$ mol. Mass $K_2O = 0.1 \, \text{mol} \times 94.2 \, \text{g/mol} = 9.42$ g.
  2. What volume (in dm³) of hydrogen gas is produced at STP when 19.5 g of zinc reacts with excess HCl? $Zn + 2HCl \rightarrow ZnCl_2 + H_2$.
    Solution: Moles Zn = $\frac{19.5}{65.4} \approx 0.298$ mol. Mole ratio Zn:Hâ‚‚ is 1:1. Moles Hâ‚‚ = 0.298 mol. Volume Hâ‚‚ at STP = $0.298 \, \text{mol} \times 22.4 \, \text{dm}^3/\text{mol} \approx 6.68 \, \text{dm}^3$.
  3. What volume (in cm³) of $0.5 \, \text{mol dm}^{-3}$ HCl is required to react completely with 1.94 g of magnesium? $Mg + 2HCl \rightarrow MgCl_2 + H_2$.
    Solution: Moles Mg = $\frac{1.94}{24.3} \approx 0.08$ mol. Mole ratio Mg:HCl is 1:2. Moles HCl = $0.08 \times 2 = 0.16$ mol. Volume HCl = $\frac{\text{moles}}{\text{conc}} = \frac{0.16}{0.5} = 0.32 \, \text{dm}^3 = 320 \, \text{cm}^3$.
  4. In a titration, $27.3 \, \text{cm}^3$ of an HCl solution was required to neutralize $25.0 \, \text{cm}^3$ of a $0.10 \, \text{mol dm}^{-3}$ NaOH solution. What was the concentration of the acid? $HCl + NaOH \rightarrow NaCl + H_2O$.
    Solution: Moles NaOH = $0.10 \times \frac{25.0}{1000} = 0.0025$ mol. Mole ratio HCl:NaOH is 1:1. Moles HCl = 0.0025 mol. Conc. HCl = $\frac{\text{moles}}{\text{volume}} = \frac{0.0025}{\frac{27.3}{1000}} \approx 0.0916 \, \text{mol dm}^{-3}$.
  5. What volume of oxygen at 298 K and 100 kPa is needed to react with 8.5 g of $H_2S$? $2H_2S + 3O_2 \rightarrow 2SO_2 + 2H_2O$.
    Solution: Moles $H_2S = \frac{8.5}{34.1} \approx 0.249$ mol. Mole ratio $H_2S:O_2$ is 2:3. Moles $O_2 = 0.249 \times \frac{3}{2} \approx 0.374$ mol. Volume $O_2 = \frac{nRT}{P} = \frac{0.374 \times 8.31 \times 298}{100000} \approx 0.00926 \, \text{m}^3 = 9.26 \, \text{dm}^3$.
  6. What mass of aluminum oxide is produced when 135 g of aluminum is burned in oxygen? $2Al + 1.5O_2 \rightarrow Al_2O_3$.
    Solution: Moles Al = $\frac{135}{27} = 5$ mol. Mole ratio Al:$Al_2O_3$ is 2:1. Moles $Al_2O_3 = 5 \times \frac{1}{2} = 2.5$ mol. Mass $Al_2O_3 = 2.5 \, \text{mol} \times 102 \, \text{g/mol} = 255$ g.
  7. What volume of CO₂ in dm³ is produced when 5.6 g of butene ($C_4H_8$) is burnt at RTP? $C_4H_8 + 6O_2 \rightarrow 4CO_2 + 4H_2O$.
    Solution: Moles $C_4H_8 = \frac{5.6}{56} = 0.1$ mol. Mole ratio $C_4H_8:CO_2$ is 1:4. Moles $CO_2 = 0.1 \times 4 = 0.4$ mol. Volume at RTP = $0.4 \, \text{mol} \times 24.4 \, \text{dm}^3/\text{mol} = 9.76 \, \text{dm}^3$.
  8. What mass of lead(II) chloride is obtained from 37.2g of $PbO_2$? $PbO_2 + 4HCl \rightarrow PbCl_2 + Cl_2 + 2H_2O$.
    Solution: Moles $PbO_2 = \frac{37.2}{239.2} \approx 0.1555$ mol. Mole ratio $PbO_2:PbCl_2$ is 1:1. Moles $PbCl_2 = 0.1555$ mol. Mass $PbCl_2 = 0.1555 \, \text{mol} \times 278.2 \, \text{g/mol} \approx 43.3$ g.
  9. What mass of magnesium oxide is formed when magnesium reacts with 6 dm³ of oxygen at RTP? $2Mg + O_2 \rightarrow 2MgO$.
    Solution: Moles Oâ‚‚ = $\frac{6}{24.4} \approx 0.246$ mol. Mole ratio Oâ‚‚:MgO is 1:2. Moles MgO = $0.246 \times 2 = 0.492$ mol. Mass MgO = $0.492 \, \text{mol} \times 40.3 \, \text{g/mol} \approx 19.8$ g.
  10. What volume of hydrogen is produced when 195 g of potassium is added to water at 298 K and 100 kPa? $2K + 2H_2O \rightarrow 2KOH + H_2$.
    Solution: Moles K = $\frac{195}{39.1} = 5$ mol. Mole ratio K:Hâ‚‚ is 2:1. Moles Hâ‚‚ = $5 \times \frac{1}{2} = 2.5$ mol. Volume Hâ‚‚ = $\frac{nRT}{P} = \frac{2.5 \times 8.31 \times 298}{100000} \approx 0.0619 \, \text{m}^3 = 61.9 \, \text{dm}^3$.

End of Topic Questions

Answer: A starting substance in a chemical reaction, written on the left side of the equation.

Answer: In a closed system, the total mass of the reactants equals the total mass of the products because atoms are conserved.

Answer: $P_4 + 5O_2 \rightarrow P_4O_{10}$.

Answer: Using the mole ratio from the balanced equation to convert moles of a given substance to moles of a required substance.

Answer: The relative number of moles of reactants and products in a reaction.

Answer: $2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O$.

Answer: The ratio of $O_2:H_2O$ is 1:2. So, $2 \times 2 = 4$ moles of $H_2O$ are produced.

Answer: Solid, liquid, gas, and aqueous (dissolved in water).

Answer: To ensure the mole ratios used for calculations are correct and reflect the Law of Conservation of Mass.

Answer: $2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O$.

Answer: 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃.

Answer: Moles Mg = 12/24.3 ≈ 0.49 mol. Moles MgO = 0.49 mol. Mass MgO = $0.49 \times 40.3 \approx 19.9$ g.

Answer: $2AgNO_3 + CaCl_2 \rightarrow 2AgCl + Ca(NO_3)_2$.

Answer: Convert the given mass of the reactant into moles.

Answer: $2K + 2H_2O \rightarrow 2KOH + H_2$.

Answer: Moles $CaCO_3$ = 100/100 = 1 mol. Moles $CO_2$ = 1 mol. Volume at STP = $1 \times 22.4 = 22.4 \, \text{dm}^3$.

Answer: The ratio is 4:4 or 1:1. So, 2 moles of NO are produced.

Answer: The calculation of quantities of reactants and products involved in a chemical reaction using mole ratios from a balanced equation.

Answer: $Al_2(SO_4)_3 + 3Ca(OH)_2 \rightarrow 2Al(OH)_3 + 3CaSO_4$.

Answer: Converting the calculated moles of the product into mass by multiplying by its molar mass.
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