JPJC 2026 Mole & Stoichiometry MCQ Answer

Jurong Pioneer Junior College 2026 JC1 H2 Chemistry: The Mole Concept and Stoichiometry 1 JURONG PIONEER JUNIOR COLLEGE 2026 JC1 H2 Chemistry (9476) MCQ Quiz Topic 2: The Mole Concept & Stoichiometry Name : Worked Solutions 5 Class : 26S________________ Date : ____________ Duration : 10 minutes Answer all questions. For each question there are four possible answers, A, B, C and D. Choose the one you consider correct and indicate your answers in the brackets provided. 1 [2022 P1 Q7] & [2019 P1 Q11] ‘Solder glass’ is composed of 16% boron oxide, B 2O3, by mass, and 84% lead(II) oxide, PbO, by mass. What is the Pb/B molar ratio in ‘solder glass’? A 0.82 B 0.91 C 1.22 D 1.64 ( ) Ans: A Assuming 100 g of ‘solder glass’, there are 16 g of B2O3 and 84 g of PbO. n(B2O3) present = 16 69.6 = 0.230 mol n(B) present = 0.230 × 2 = 0.460 mol (since mole ratio of B : B2O3 = 2 : 1) n(PbO) present = 84 223.2 = 0.376 mol n(Pb) present = = 0.376 mol (since mol ratio of Pb : PbO = 1 : 1) molar ratio of Pb B = 0.376 0.460 = 0.82.

2 2 [2016 P1 Q2] Analysis of a mixture of two sulfur–containing gases shows that hydrogen sulfide, H2S, and carbon disulfide, CS2, are present in a 2 : 1 mole ratio. This mixture is burned in excess oxygen. What will the SO2 : CO2 mole ratio in the mixture obtained after complete combustion? A 1 : 1 B 2 : 1 C 3 : 1 D 4 : 1 ( ) Ans: D Let the amount of CS2 present be x mol. CS2 + 3O2 → CO2 + 2SO2 x mol x mol 2x mol reacted formed H2S + 1.5O2 → H2O + SO2 2x mol 2x mol reacted formed Total amount of SO2 formed = 2x + 2x = 4x mol Amount of SO2 formed = x mol  mole ratio of SO2 : CO2 = 4 : 1 Given H2X : CS2 2 : 1

Jurong Pioneer Junior College 2026 JC1 H2 Chemistry: The Mole Concept and Stoichiometry 3 3 10 cm3 of a gaseous hydrocarbon X was exploded in excess oxygen. Upon cooling to room temperature, there was a contraction in volume of 35 cm3. When the products were treated with excess KOH(aq), there was a further contraction of 40 cm3. What is the molecular formula of hydrocarbon X? [All gaseous volumes are measured under identical conditions at r.t.p.] A C2H4 B C4H8 C C4H10 D C4H12 ( ) Ans: C Second contraction = Volume of CO2 = 40 cm3 First contraction = x y 2 2C H CO O reactedV + V - V 35 = 10 + 2O reactedV - 40 Volume of O2 reacted = 65 cm3 CxHy(g) + (x + y 4 ) O2(g) → x CO2(g) + y 2 H2O(l) at r.t.p. Volume/cm3 10 65 40 - mole ratio 1 : 6.5 : 4 1 : (x + y 4 ) : x  x = 4 and (4 + y 4 ) = 6.5  y = 10 Alternatively, CxHy(g) + (x + y 4 ) O2(g) → x CO2(g) + y 2 H2O(l) at r.t.p. 10 cm3 10(x + y 4 ) cm3 10x cm3 - (negligible) First contraction = x y 2 2C H CO O reactedV + V - V 35 = 10 + 10(x + y 4 ) - 10x 35 = 10 + 10x + 2.5y - 10x 2.5y = 25  y = 10 Second contraction = Volume of CO2 = 40 cm3 10x = 40  x = 4  the molecular formula of the hydrocarbon is C4H10.

4 4 The manufacture of calcium carbide involves the following reaction: CaO(s) + 3C(s) → CaC2(s) + CO(g) What is the maximum mass of calcium carbide (in kg) that can be obtained from 40.0 kg of calcium oxide and 40.0 kg of coke (carbon)? A 13.3 B 40.0 C 45.7 D 71.2 ( ) Ans: C n(CaO) available = 40 000 56.1 = 713 mol ✓ (LR) n(coke) = 40 000 12.0 = 3333 mol CaO : C 1 : 3 713 mol of CaO will react with (3  713 =) 2139 mol of coke for complete reaction.  CaO is the limiting reagent. CaO : CaC2 1 : 1 713 : ? maximum n(CaC2) produced = 713 mol  maximum mass of CaC2 = 713 64.1 = 45 703 g = 45.7 kg

Jurong Pioneer Junior College 2026 JC1 H2 Chemistry: The Mole Concept and Stoichiometry 5 5 25.0 cm3 of 0.05 mol dm−3 KClO4(aq) required 50.0 cm3 of 0.20 mol dm−3 TiCl3 (aq) to reach the end–point in a titration. Given that titanium(III) is oxidised to titanium(IV) in this reaction, which one of the following formulae correctly represents the reduction product of the ClO4− ion? A ClO− B Cl− C ClO2− D ClO3− ( ) Ans: B Given: [O] Ti3+ → Ti4+ + e– n(Ti3+) = 0.20  50.0 1000 = 0.01 mol = n(e-) transferred n(ClO4-) = 0.05  25.0 1000 = 0.00125 mol ClO4- : e– 0.00125 : 0.01 1 : 8  ClO4- undergoes reduction and oxidation no. of Cl decreases by 8 from +7 to -1. ClO– Cl- ClO2- ClO3- (✓) (+7) (+1) (-1) (+3) (+5)