Level 1 — RecognitionChemical Reactions & Stoichiometry

Chemical Reactions & Stoichiometry

30 marksprintable — key stays hidden on paper

Level 1 Test — Recognition

Time: 20 minutes Total Marks: 30

Section A — Multiple Choice (1 mark each) — 10 marks

Choose the single best answer.

Q1. When the equation $\text{H}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O}$ is balanced with smallest whole-number coefficients, the coefficient of $\text{O}_2$ is: (a) 1 (b) 2 (c) $\tfrac{1}{2}$ (d) 3

Q2. The reaction $2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2$ is an example of: (a) Combination (b) Decomposition (c) Displacement (d) Double displacement

Q3. The oxidation number of sulfur in $\text{H}_2\text{SO}_4$ is: (a) $+2$ (b) $+4$ (c) $+6$ (d) $-2$

Q4. Reaction of $\text{AgNO}_3(aq) + \text{NaCl}(aq) \rightarrow \text{AgCl}(s) + \text{NaNO}_3(aq)$ is classified as: (a) Combination (b) Redox (c) Double displacement (d) Combustion

Q5. In the reaction $\text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu}$ , the species oxidised is: (a) $\text{Cu}^{2+}$ (b) $\text{Zn}$ (c) $\text{SO}_4^{2-}$ (d) $\text{Cu}$

Q6. Percent yield is defined as: (a) $\dfrac{\text{theoretical}}{\text{actual}}\times 100$ (b) $\dfrac{\text{actual}}{\text{theoretical}}\times 100$ (c) actual $-$ theoretical (d) $\dfrac{\text{actual}}{100}$

Q7. How many moles of solute are in $250\ \text{mL}$ of $0.40\ \text{M}$ HCl? (a) 0.10 (b) 0.040 (c) 1.6 (d) 0.25

Q8. In $\text{KMnO}_4$ , the oxidation number of Mn is: (a) $+2$ (b) $+4$ (c) $+7$ (d) $+6$

Q9. Complete combustion of $\text{CH}_4$ produces: (a) $\text{CO} + \text{H}_2$ (b) $\text{CO}_2 + \text{H}_2\text{O}$ (c) C + $\text{H}_2\text{O}$ (d) $\text{CO}_2 + \text{H}_2$

Q10. In a limiting reagent problem, the limiting reagent is the one that: (a) is present in largest mass (b) is fully consumed first (c) remains in excess (d) has largest molar mass

Section B — Matching (1 mark each) — 5 marks

Q11. Match each reaction in Column X to its type in Column Y.

Column X	Column Y
(i) $\text{CaO} + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2$	(A) Displacement
(ii) $\text{Fe} + \text{CuSO}_4 \rightarrow \text{FeSO}_4 + \text{Cu}$	(B) Neutralization
(iii) $2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2$	(C) Combination
(iv) $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$	(D) Combustion
(v) $\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$	(E) Decomposition

Section C — True/False WITH Justification (3 marks each: 1 T/F + 2 justification) — 15 marks

Q12. The oxidation number of oxygen is always $-2$ in every compound.

Q13. In a redox reaction, the oxidising agent is itself reduced.

Q14. Diluting a solution changes the number of moles of solute present.

Q15. In the reaction $\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$ , the salt formed is $\text{Na}_2\text{SO}_4$ .

Q16. A balanced chemical equation must have equal numbers of atoms of each element on both sides, satisfying conservation of mass.

Answer keyMark scheme & solutions

Section A (1 mark each)

Q1. (a) 1 — Balanced: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ ; smallest whole numbers give $\text{O}_2$ coefficient = 1. (1)

Q2. (b) Decomposition — one reactant breaks into two products. (1)

Q3. (c) $+6$ — H is $+1$ (total $+2$ ), O is $-2$ (total $-8$ ); $S + 2 - 8 = 0 \Rightarrow S = +6$ . (1)

Q4. (c) Double displacement — ions exchange partners; precipitate AgCl forms. (1)

Q5. (b) Zn — Zn goes $0 \to +2$ (loses electrons = oxidised). (1)

Q6. (b) $\dfrac{\text{actual}}{\text{theoretical}}\times 100$ . (1)

Q7. (a) 0.10 — $n = M \times V = 0.40 \times 0.250 = 0.10\ \text{mol}$ . (1)

Q8. (c) $+7$ — K is $+1$ , O total $-8$ ; $Mn + 1 - 8 = 0 \Rightarrow Mn = +7$ . (1)

Q9. (b) $\text{CO}_2 + \text{H}_2\text{O}$ — complete combustion of hydrocarbon. (1)

Q10. (b) is fully consumed first — limits the amount of product. (1)

Section B

Q11. (5 marks, 1 each) (i) → C (combination); (ii) → A (displacement); (iii) → E (decomposition); (iv) → B (neutralization); (v) → D (combustion). (5)

Section C (1 mark T/F + 2 marks justification)

Q12. FALSE. (1) Justification: Oxygen is usually $-2$ , but exceptions exist: in peroxides (e.g. $\text{H}_2\text{O}_2$ ) it is $-1$ , and in $\text{OF}_2$ it is $+2$ . (2)

Q13. TRUE. (1) Justification: The oxidising agent accepts electrons from another species; gaining electrons means its oxidation number decreases, i.e. it is reduced. (2)

Q14. FALSE. (1) Justification: Dilution adds solvent only; moles of solute stay constant ( $n = M_1V_1 = M_2V_2$ ). Only concentration (mol/L) decreases. (2)

Q15. TRUE. (1) Justification: Neutralization: the cation of the base ( $\text{Na}^+$ ) combines with the anion of the acid ( $\text{SO}_4^{2-}$ ) to form salt $\text{Na}_2\text{SO}_4$ ; water is the other product. (2)

Q16. TRUE. (1) Justification: A balanced equation reflects the law of conservation of mass — atoms are neither created nor destroyed, so each element's atom count must be equal on both sides. (2)

[
  {"claim":"Q3: S oxidation number in H2SO4 is +6","code":"H=1; O=-2; S=symbols('S'); sol=solve(Eq(2*H+S+4*O,0),S); result=(sol[0]==6)"},
  {"claim":"Q7: moles = 0.40 M * 0.250 L = 0.10 mol","code":"n=Rational(40,100)*Rational(250,1000); result=(n==Rational(1,10))"},
  {"claim":"Q8: Mn oxidation number in KMnO4 is +7","code":"K=1; O=-2; Mn=symbols('Mn'); sol=solve(Eq(K+Mn+4*O,0),Mn); result=(sol[0]==7)"},
  {"claim":"Q1: balancing H2+O2->H2O gives O2 coefficient 1 with 2H2 and 2H2O","code":"a,b,c=symbols('a b c'); sol=solve([Eq(2*a,2*c),Eq(2*b,c)],[a,c],dict=True); s=sol[0]; result=(s[a].subs(b,1)==2 and s[c].subs(b,1)==2)"}
]