Level 2 — RecallChemical Reactions & Stoichiometry

Chemical Reactions & Stoichiometry

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Level 2 (Recall / Standard Problems)

Time: 30 minutes | Total Marks: 40

Use molar masses: H = 1, C = 12, N = 14, O = 16, Na = 23, S = 32, Cl = 35.5, Fe = 56, Zn = 65, Ca = 40, K = 39.

Q1. Balance the following chemical equations. (4 marks, 1 each) (a) $Fe + O_2 \rightarrow Fe_2O_3$ (b) $C_3H_8 + O_2 \rightarrow CO_2 + H_2O$ (c) $Al + HCl \rightarrow AlCl_3 + H_2$ (d) $KClO_3 \rightarrow KCl + O_2$

Q2. Classify each of the following reactions as combination, decomposition, displacement, or double displacement. (4 marks, 1 each) (a) $CaO + H_2O \rightarrow Ca(OH)_2$ (b) $Zn + CuSO_4 \rightarrow ZnSO_4 + Cu$ (c) $AgNO_3 + NaCl \rightarrow AgCl + NaNO_3$ (d) $2Pb(NO_3)_2 \rightarrow 2PbO + 4NO_2 + O_2$

Q3. Assign the oxidation number of the underlined atom. (4 marks, 1 each) (a) $\underline{S}$ in $H_2SO_4$ (b) $\underline{Mn}$ in $KMnO_4$ (c) $\underline{Cr}$ in $Cr_2O_7^{2-}$ (d) $\underline{N}$ in $NH_4^+$

Q4. For the reaction $N_2 + 3H_2 \rightarrow 2NH_3$ , if $28\ \text{g}$ of $N_2$ reacts with $10\ \text{g}$ of $H_2$ : (5 marks) (a) Identify the limiting reagent. (3) (b) Calculate the mass of $NH_3$ formed. (2)

Q5. Define theoretical yield and percent yield. In a reaction the theoretical yield of a product is $50\ \text{g}$ but only $42\ \text{g}$ is obtained. Calculate the percent yield. (4 marks)

Q6. $25.0\ \text{mL}$ of $HCl$ is neutralized by $20.0\ \text{mL}$ of $0.100\ \text{M}$ $NaOH$ . (4 marks) (a) Write the neutralization equation. (1) (b) Calculate the molarity of the $HCl$ solution. (3)

Q7. Balance the following redox reaction in acidic medium by the ion-electron (half-reaction) method: (5 marks) $Fe^{2+} + MnO_4^- \rightarrow Fe^{3+} + Mn^{2+}$

Q8. How many mL of water must be added to $50\ \text{mL}$ of $2.0\ \text{M}$ $H_2SO_4$ to make the solution $0.5\ \text{M}$ ? (3 marks)

Q9. Write the balanced equation for the complete combustion of ethanol $C_2H_5OH$ , and calculate the mass of $CO_2$ produced when $23\ \text{g}$ of ethanol is burned completely. (4 marks)

Q10. State whether each is oxidation or reduction, based on oxidation-number change: (3 marks, 1 each) (a) $Zn \rightarrow Zn^{2+}$ (b) $Cl_2 \rightarrow 2Cl^-$ (c) $Fe^{2+} \rightarrow Fe^{3+}$

Answer keyMark scheme & solutions

Q1. (4) (a) $4Fe + 3O_2 \rightarrow 2Fe_2O_3$ — balances Fe (4) and O (6). ✓ (1) (b) $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$ — C:3, H:8, O:10. ✓ (1) (c) $2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2$ — Al:2, Cl:6, H:6. ✓ (1) (d) $2KClO_3 \rightarrow 2KCl + 3O_2$ — K:2, Cl:2, O:6. ✓ (1)

Q2. (4) (a) Combination (two reactants → one product). (1) (b) Displacement (Zn displaces Cu). (1) (c) Double displacement (ion exchange, precipitate). (1) (d) Decomposition (one reactant → several products). (1)

Q3. (4) Using rules O = −2, H = +1, K = +1, net charge conservation. (a) $2(+1)+S+4(-2)=0 \Rightarrow S=+6$ . (1) (b) $+1+Mn+4(-2)=0 \Rightarrow Mn=+7$ . (1) (c) $2Cr+7(-2)=-2 \Rightarrow 2Cr=+12 \Rightarrow Cr=+6$ . (1) (d) $N+4(+1)=+1 \Rightarrow N=-3$ . (1)

Q4. (5) Moles: $N_2 = 28/28 = 1\ \text{mol}$ ; $H_2 = 10/2 = 5\ \text{mol}$ . (1) Required ratio $N_2:H_2 = 1:3$ . For 1 mol $N_2$ need 3 mol $H_2$ ; have 5 mol → $H_2$ in excess. (1) Limiting reagent = $N_2$ . (1) $NH_3$ formed $= 2 \times 1 = 2\ \text{mol}$ ; mass $= 2 \times 17 = 34\ \text{g}$ . (2)

Q5. (4) Theoretical yield: maximum product mass predicted by stoichiometry assuming complete reaction of limiting reagent. (1) Percent yield $= \dfrac{\text{actual}}{\text{theoretical}}\times 100$ . (1) $= \dfrac{42}{50}\times100 = 84\%$ . (2)

Q6. (4) (a) $HCl + NaOH \rightarrow NaCl + H_2O$ . (1) (b) mol NaOH $= 0.100 \times 0.0200 = 0.00200$ mol. (1) 1:1 ratio → mol HCl $= 0.00200$ . (1) $M_{HCl} = 0.00200/0.0250 = 0.080\ \text{M}$ . (1)

Q7. (5) Oxidation: $Fe^{2+} \rightarrow Fe^{3+} + e^-$ . (1) Reduction: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ . (2) Multiply oxidation by 5 to balance electrons; add: $5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$ (2) (Charge: LHS $+17$ , RHS $+17$ ✓)

Q8. (3) $M_1V_1 = M_2V_2$ : $2.0\times50 = 0.5\times V_2 \Rightarrow V_2 = 200\ \text{mL}$ . (2) Water added $= 200 - 50 = 150\ \text{mL}$ . (1)

Q9. (4) $C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O$ . (2) Moles ethanol $= 23/46 = 0.5$ mol. (1) $CO_2 = 2\times0.5 = 1$ mol $= 44\ \text{g}$ . (1)

Q10. (3) (a) Oxidation (0 → +2, loses electrons). (1) (b) Reduction (0 → −1, gains electrons). (1) (c) Oxidation (+2 → +3). (1)

[
  {"claim":"Q4: N2 limiting, NH3 mass = 34 g","code":"nN2=28/28; nH2=10/2; nNH3=2*min(nN2,nH2/3); mass=nNH3*17; result=(mass==34)"},
  {"claim":"Q5: percent yield = 84","code":"py=42/50*100; result=(py==84)"},
  {"claim":"Q6: HCl molarity = 0.080 M","code":"molNaOH=0.100*0.0200; M=molNaOH/0.0250; result=(abs(M-0.080)<1e-9)"},
  {"claim":"Q8: water added = 150 mL","code":"V2=2.0*50/0.5; water=V2-50; result=(water==150)"},
  {"claim":"Q9: CO2 mass = 44 g from 23 g ethanol","code":"nEt=23/46; nCO2=2*nEt; mass=nCO2*44; result=(mass==44)"}
]