Question

CHEMIOS CHAPTERS 34 EXAMA 9. Complete combustion of a 0.70-mal sample of a hydrocarbon. C gives 3.50 mal of C0, and 2.80 mol of H.O. The molecular formula of the original hydrocarbon is a. Cal h. Ch. c. Cli, d.Cls. &. C.H. 10. Aspirin is produced by the reaction of cylicid =138.1 mol and acetic anhydride (N=102.1 ml C-H40, 5) +C ,0 4,ỄN - CH 40 (8) + C3H40, 4) If 2.04 g of C O - 189.2 g/mol) is produced from the reaction of 303 gC,H,O, and 4.01 8C,H,O, what is the percent yield? a. 28.8% b. 29.0% 50.04 510 673% 11. What is the oxidation number of each om in sodium Indron carb Nah a Na +1.1--10- 6, 0- 2 b. Na + H .C- 0 --2 c. Na - 1, - 1.C- +2,0 2 d. Na-1. 1.0-0.0- 3 c. Na-0,000.00 12. What base results when K, reacts with water a. 11,0 (a) la o cad. Ko c .KO 13. The commercial production of phosphoric acid, H,PO can be represented by the equation 1500g 300g 3048 1190 E 300 Ca(PO4 + 3SO, + SC+ SO, N O CASO, +500, +2H.PO 310 ml 60 ml 12.0 gmel 120 18.0 ml The molar mass for each reactant is shown on the rantand the ma c h rectant for this problem is given above. Which substance is the limiting reactant H.O bc 4.0, d. C PO eso 14. What are the spectations in the reaction betwe e n acid and ammonia? 2. Honly b. NO, only c. II and NHL d. No," and NH, e. H. NO, and NH, 15. Which species is reduced in the reaction below? (2)+CIO (4) -10 + (0) ar H.O cad.10 . Cosmote Come convey Powered by Copere

Answer 1

gr. Givena Hydrocarbon is atly Cally to xcQ + H2O Camy = a mole of co₂ formed and y mole of JAN M20 formed Griven 0.70 mole of Gly co formed = reached 0.70* 420 formed = 0.70 y Alq + 0.704 = 3.50 Alge 0.70 y = 2.80 25 y=8 Hence, Hydrocarbon = (54g (b) option is correct. 107 GHEOz (s) + CQH6Old) (94804 Is) + (2 6 HQ 02 (2) given Googl/= 3.03g (4460318) -4.01g from a limiting reagent is GHG Oz (s). From Imole of CHO = Imole of (gHq 04 Formed moles of GHQO3 = 3.03 mole 138.1 Hence; 3.03 138.1 mole of Cg H804 formed, The critical Cg Hg Oq formed = 3.03 138-1 x Molar mass of Cg Hoon = 3.03 x 180.2 g 138.1 = 3.959 gielda Actual Thebr}\વા X1oo. yield = 51.6% option-d

117 oxidation no. of Nallco Na=t1; (=+4; 0=-2 ; let ubex +1+&+4-6=0 Wa=t1; H+1; (= +4; 0=-2 option - b K2O + H2O → zkon lap) Base formed = koulag (di option. 147 reaction between aqueous HNO3 and ammonia NH₃ 6919) + H Nolaq! NH4NO3lag) Ionic, NHq1g) + ut +NO3 lag! NHA Tlaq) + No, lagu spectator Ion = NO only I laq) + ao lagi 10 lag) tu lag reduction gain of electron in Hence; reduced 4o. species = co (e) (a (PO4 + 35102 +50 +502 +3160 Blasiu4 + 500+ 24 pon 1500g 300g 3049 1150g 300g moles of (a (Po4) = 1500 ; moles of (= 304 - 310 moles at SiO2 = 300 i moles of a = 1150 60 i moles of How = 30/18 32

limiting reagent = minimum mole from Stoichiometric wefficient. Ca, P4), - 500 ( XS S102 = 300 60X3 O2 = 1150 3285 M. d o 200 H2O=300 1883 Hence; SiO2 is limiting reagent. optim (e) is correct
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