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Question → Hess Law - g state that ita reaction can take place by more them one route and initial and final condition are the same, the total enthalpy change is the same. A > B Reactant Produc î Intermediate And the reactions are treated as treated as Algebrically In the given equation Cao + CO2 (g) Cacos Anel given thermochemical equation 2 (a + Og → DHE ? are acao DH = -1270.2 KJ cit og COL IDH = -393.5KJ To 2 latac 4302 2. Cacos , DH = -2413.8ko ges the clesived equation were treat given equation and energy Algebrically. And abbely forowing equation. aberation & - 6 - 2 weget Cate +2 / 62-4-02-46 = 26 Calos-co2- Cao DHE Àx-243.8 - (-39315)-(-127-2)}} Cao tlog Calez OH= -1206.9 +393. $ +6351 k 0143 -1206.9+1028.6 RJ DH = 178.3 KJ
Question-a. Given, mass of samples 2010 ovog mass of hun sample 1:34g 8.00g of oft in Compound= massofu mass of sample .X100 1:34g 20:00 x60 = 6.7% & ofc in compound= mass of c. xoo = 800g, xwo mass of sample 20:09 = 40% % of o in Compound = 100 [y of ctx of us forowing operation whole Number Ratio 100 -(40+607) 100 - 46.7 - 53:3% To Calculate empirical formela we use element, Atomic Relatie moles Sisablest % mass Ratio Atomicmas с 12 40 40 13:33 = 3:33 13:33 16 5313 13:33 23:33 76 H 67 6:7 67 2. 3.33 Hence, empirical formula Go, H₂ S33 3:33 = 1 - 6:7 2
Question-3 Given reaction is Q NAOH + Cog-> & moi Na coz +H2O I ning! imel innes / Given, moles of naona 3070 mo? 11 CO₂ = 9.00 mol amol Naou react with 1 mol of CO2 3.70 moi Imel x3,70 mer a moi from Stoichiometry of Reaction. of of of con 1.85 mor of con But we have 2.00 mol of con. Hence, con remain in excess reagent and Naok is limiting reagent noles of excess reactant remain - (2.00 -1185) mai of ence and is excess Thus, Initial - Consumed 0.15 Mei Thus, 0.15 mol excess reactant remains.