RI 2024 Mole Concept and Stoichiometry Tutorial (Ans)

-1- Raffles Institution Year 5 H2 Chemistry 2024 Tutorial 1 – The Mole Concept and Stoichiometry (Suggested Answers) Answers 1 (a) Mass of F– in 1000 cm3 of fluorinated reservoir water = 10  0.16 = 1.6 mg Optional working: Mass of F– in 100 cm3 of diluted solution = 0.1  0.16 = 0.016 mg Mass of F– in 10.0 cm3 of fluorinated reservoir water = 0.016 mg Mass of F– in 1000 cm3 of fluorinated reservoir water = 100  0.016 = 1.60 mg (b) Amount of F– in fluorinated reservoir water = 1.6×10−3 19.0 = 8.421  10–5 mol Amount of H2SiF6 required = 1 6 × 8.421  10-5 = 1.404  10–5 mol Mass of H2SiF6 required = (1.404  10-5)(144.1) = 2.02  10–3 g = 2.02 mg (Note: answer can be given in grams or milligrams, since the question did not specify.) 2 3 23 3 5.30Concentration of CO 10.6 g dm500 10 − −== M M2CO3 + 2HCl → 2MCl + H2O + CO2 326.0Amount of HC 0.192 4.992 10 mol1000 −=  = l 33 23 1Amount of CO 4.992 10 2.496 10 mol2 −−=   = M 3 3 23 3 2.496 10Concentration of CO 0.09984 mol dm25.0 10 − − − == M Let the relative atomic mass of M be x. ( )0.09984 2 12.0 3 16.0 10.6 23.1 x x + +  = = 3 (a) C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) N2(g) + 2O2(g) → 2NO2(g) (b) Applying Avogadro’s law, V is proportional to n, Volume of CO2 formed = 10 × 3 = 30 cm3 (c) 50Amount of NaOH present initially 0.50 0.02 5 mol1000=  = 2NaOH + H2SO4 → Na2SO4 + 2H2O 3 24 24.40Amount of H SO used 0.35 8.54 10 mol1000 −=  =  3 24Amount of NaOH reacted with H SO 2 8.54 10 0. 01708 mol−=   = 3Amount of NaOH reacted with gaseous prod ucts 0.025 0.01708 7.92 10 mol −= − = 

-2- (d) CO2(g) + 2NaOH(aq) → Na2CO3(aq) + H2O(l) At room temperature and pressure, 3 3 2 30 10Amount of NaOH reacted with CO 2 2.50 10 mo l24 − −=  =  3 3 3 2Amount of NaOH reacted with NO 7.92 10 2.50 10 5.42 10 mol− − −=  −  =  Since mole ratio of NO2 to NaOH is 1:1 (as given in the equation), 3 2 3 3 Amount of NO 5.42 10 mol 2 10 5.42 1024 65.0 z z − − − =  = = 4 (a) (NH4)2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2NH3(g) + 2H2O(l) HCl(aq) + NH3(g) → NH4Cl(aq) Or NH4+(aq) + OH–(aq) → NH3(g) + H2O(l) H+(aq) + NH3(g) → NH4+(aq) (b) NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) 3 Amount of unreacted HC (step III) Amount of NaOH used 20.0 0.100 2.00 10 mol1000 − = =  =  l 350.0Amount of HC (step II) 0.100 5.00 10 mol1000 −=  = l 33 33 3 Amount of NH Amount of HC reacted with l iberated NH 5.00 10 2.00 10 3.00 10 mol Amount of N present in 1.00 g of a foodstuff −− − = =  −  =  = l 3Mass of N 3.00 10 14.0 0.0420 g−=   = 0.0420% by mass of N in the foodstuff 100% 4.20%1.00=  =