Level 4 — ApplicationChemical Reactions & Stoichiometry

Chemical Reactions & Stoichiometry

⏱ 60 minutes50 marksprintable — key stays hidden on paper

Level 4 — Application (novel problems, no hints) Time limit: 60 minutes Total marks: 50

Useful data: Molar masses (g/mol): H = 1.0, C = 12.0, N = 14.0, O = 16.0, Na = 23.0, S = 32.0, Cl = 35.5, K = 39.0, Ca = 40.0, Mn = 55.0, Fe = 56.0, Cu = 63.5, Ag = 108.0. Molar gas volume at STP = 22.4 L/mol.

Question 1 — Limiting reagent & percent yield [12 marks]

Ammonia is produced industrially by the Haber process: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$

A reactor is charged with $28.0\ \text{kg}$ of $N_2$ and $7.5\ \text{kg}$ of $H_2$ .

(a) Determine the limiting reagent, showing your reasoning. (4) (b) Calculate the theoretical yield of $NH_3$ in kg. (3) (c) The reactor actually produces $12.75\ \text{kg}$ of $NH_3$ . Calculate the percent yield. (2) (d) Calculate the mass (kg) of the excess reagent that remains unreacted. (3)

Question 2 — Redox balancing (ion-electron method) [12 marks]

In acidic solution, permanganate ion oxidises oxalate ion: $MnO_4^- + C_2O_4^{2-} \rightarrow Mn^{2+} + CO_2$

(a) Assign the oxidation number of carbon in $C_2O_4^{2-}$ and in $CO_2$ , and state the change per carbon atom. (3) (b) Balance the equation completely using the ion-electron (half-reaction) method in acidic medium. Show both half-reactions. (7) (c) Identify the oxidising agent and the reducing agent. (2)

Question 3 — Titration / solution stoichiometry [12 marks]

A $25.00\ \text{mL}$ sample of household vinegar (aqueous acetic acid, $CH_3COOH$ ) is titrated with $0.0500\ \text{mol/L}$ NaOH. The equivalence point requires $34.60\ \text{mL}$ of the NaOH solution.

(a) Write the balanced neutralisation equation and name the salt formed. (2) (b) Calculate the molar concentration of acetic acid in the vinegar. (3) (c) Express this concentration as a mass/volume percent (g of $CH_3COOH$ per $100\ \text{mL}$ ), assuming solution density $\approx 1.00\ \text{g/mL}$ . (3) (d) The vinegar was actually a diluted sample: $10.00\ \text{mL}$ of the original vinegar had been diluted to $250.0\ \text{mL}$ before the titration. Find the acetic acid concentration in the original vinegar. (4)

Question 4 — Combustion stoichiometry [10 marks]

A gaseous hydrocarbon fuel is burned completely in oxygen. Complete combustion of $0.100\ \text{mol}$ of the fuel produces $13.2\ \text{g}$ of $CO_2$ and $5.4\ \text{g}$ of $H_2O$ .

(a) Determine the molecular formula of the hydrocarbon. (4) (b) Write the balanced combustion equation. (2) (c) Calculate the volume of $O_2$ at STP required to burn $0.100\ \text{mol}$ of the fuel. (2) (d) Classify this reaction under two of the reaction types studied and justify each classification. (2)

Question 5 — Mixed reasoning [4 marks]

$50.0\ \text{mL}$ of $0.200\ \text{mol/L}$ $AgNO_3$ is mixed with $50.0\ \text{mL}$ of $0.150\ \text{mol/L}$ $CaCl_2$ .

(a) Write the balanced net ionic equation and classify the reaction type. (2) (b) Calculate the mass of precipitate formed, identifying the limiting reagent. (2)

Answer keyMark scheme & solutions

Question 1 (12)

(a) Limiting reagent (4)

Moles $N_2 = 28000/28.0 = 1000\ \text{mol}$ . (1)
Moles $H_2 = 7500/2.0 = 3750\ \text{mol}$ . (1)
$N_2$ needs $3\times1000 = 3000\ \text{mol}\ H_2$ ; available $3750 > 3000$ , so $H_2$ is in excess. (1)
Limiting reagent = $N_2$ . (Ratio check: $H_2$ would need $1250\ \text{mol}\ N_2$ but only $1000$ present, consistent.) (1)

(b) Theoretical yield (3)

$NH_3 = 2\times n(N_2) = 2\times1000 = 2000\ \text{mol}$ . (1)
$M(NH_3) = 17.0\ \text{g/mol}$ . (1)
Mass $= 2000\times17.0 = 34000\ \text{g} = 34.0\ \text{kg}$ . (1)

(c) Percent yield (2)

$\%\text{yield} = \dfrac{12.75}{34.0}\times100 = 37.5\%$ . (2)

(d) Excess remaining (3)

$H_2$ consumed $= 3000\ \text{mol}$ . (1)
$H_2$ left $= 3750-3000 = 750\ \text{mol}$ . (1)
Mass $= 750\times2.0 = 1500\ \text{g} = 1.5\ \text{kg}$ . (1)

Question 2 (12)

(a) Oxidation numbers (3)

In $C_2O_4^{2-}$ : O = $-2$ , so $2C + 4(-2) = -2 \Rightarrow C = +3$ . (1)
In $CO_2$ : $C = +4$ . (1)
Change per C atom $= +3 \to +4$ , i.e. loss of 1 electron (oxidation). (1)

(b) Balanced equation (7)

Reduction half: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ . (2)
Oxidation half: $C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$ . (2)
Multiply: reduction ×2, oxidation ×5 to balance 10 e⁻. (1)
Add: $2MnO_4^- + 16H^+ + 5C_2O_4^{2-} \rightarrow 2Mn^{2+} + 8H_2O + 10CO_2$ (2) (Charge check: LHS $= 2(-1)+16(+1)+5(-2) = +4$ ; RHS $= 2(+2) = +4$ . ✓)

(c) Agents (2)

Oxidising agent = $MnO_4^-$ (Mn: $+7\to+2$ ). (1)
Reducing agent = $C_2O_4^{2-}$ (C: $+3\to+4$ ). (1)

Question 3 (12)

(a) Equation + salt (2) $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$ (1) Salt = sodium acetate (sodium ethanoate). (1)

(b) Concentration (3)

Moles NaOH $= 0.0500\times0.03460 = 1.730\times10^{-3}\ \text{mol}$ . (1)
1:1 ratio ⇒ moles $CH_3COOH = 1.730\times10^{-3}\ \text{mol}$ . (1)
$c = 1.730\times10^{-3}/0.02500 = 0.0692\ \text{mol/L}$ . (1)

(c) Mass/volume % (3)

$M(CH_3COOH) = 60.0\ \text{g/mol}$ . (1)
Mass per litre $= 0.0692\times60.0 = 4.152\ \text{g/L}$ . (1)
Per 100 mL $= 0.415\ \text{g}$ ⇒ 0.415% (m/v). (1)

(d) Original concentration (4)

Dilution factor $= 250.0/10.00 = 25$ . (2)
Original $c = 0.0692\times25 = 1.73\ \text{mol/L}$ . (2)

Question 4 (10)

(a) Molecular formula (4)

Moles $CO_2 = 13.2/44.0 = 0.300\ \text{mol}$ ⇒ C per fuel $= 0.300/0.100 = 3$ . (1)
Moles $H_2O = 5.4/18.0 = 0.300\ \text{mol}$ ⇒ H $= (0.300\times2)/0.100 = 6$ . (1)
Formula $= C_3H_6$ . (1)
(Check: $M = 42$ ; consistent hydrocarbon, e.g. propene/cyclopropane.) (1)

(b) Balanced combustion (2) $2C_3H_6 + 9O_2 \rightarrow 6CO_2 + 6H_2O$ (2) (or $C_3H_6 + \tfrac{9}{2}O_2 \rightarrow 3CO_2 + 3H_2O$ )

(c) Volume of $O_2$ (2)

Per mole fuel: $4.5\ \text{mol}\ O_2$ ⇒ for 0.100 mol: $0.450\ \text{mol}$ . (1)
$V = 0.450\times22.4 = 10.08\ \text{L} \approx 10.1\ \text{L}$ . (1)

(d) Classification (2)

Combustion — fuel reacts with $O_2$ producing $CO_2 + H_2O$ with heat. (1)
Redox — C is oxidised ( $-2$ in $C_3H_6 \to +4$ ), O is reduced ( $0\to-2$ ). (1)

Question 5 (4)

(a) Net ionic + type (2) $Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)$ (1) Type: double displacement / precipitation (also acceptable: metathesis). (1)

(b) Mass precipitate (2)

Moles $Ag^+ = 0.0500\times0.200 = 0.0100\ \text{mol}$ .
Moles $Cl^- = 0.0500\times0.150\times2 = 0.0150\ \text{mol}$ .
$Ag^+$ limiting (needs equal $Cl^-$ ; 0.0100 < 0.0150). Precipitate $= 0.0100\ \text{mol}\ AgCl$ . (1)
Mass $= 0.0100\times143.5 = 1.435\ \text{g} \approx 1.44\ \text{g}$ . (1)

[
{"claim":"Q1b theoretical NH3 mass = 34.0 kg","code":"nN2=28000/28.0; nNH3=2*nN2; mass=nNH3*17.0/1000; result = abs(mass-34.0)<1e-6"},
{"claim":"Q1c percent yield = 37.5%","code":"py=12.75/34.0*100; result = abs(py-37.5)<1e-3"},
{"claim":"Q1d excess H2 remaining = 1.5 kg","code":"nH2=7500/2.0; used=3*(28000/28.0); left=(nH2-used)*2.0/1000; result = abs(left-1.5)<1e-6"},
{"claim":"Q2 redox charge balance +4 = +4","code":"lhs=2*(-1)+16*(1)+5*(-2); rhs=2*(2)+8*0+10*0; result = lhs==rhs and lhs==4"},
{"claim":"Q3d original acetic acid conc = 1.73 mol/L","code":"n=0.0500*0.03460; c=n/0.02500; orig=c*(250.0/10.00); result = abs(orig-1.73)<0.01"},
{"claim":"Q4 formula C3H6 and O2 volume 10.08 L","code":"C=(13.2/44.0)/0.100; H=2*(5.4/18.0)/0.100; V=0.100*4.5*22.4; result = C==3 and H==6 and abs(V-10.08)<1e-6"},
{"claim":"Q5 AgCl mass = 1.435 g","code":"nAg=0.0500*0.200; nCl=0.0500*0.150*2; nppt=min(nAg,nCl); m=nppt*143.5; result = abs(m-1.435)<1e-3"}
]