Esercizi in preparazione all’esame di Chimica generale

BILANCIARE FORMALE REALIONI IN32 2+4+ + + 3MiSOn3H10+2HNOs3SnS2 3H,SnOs +2NOa) + - ++3 2+H- NONO5 24202 ++3e ++4+2 2Sn+3H,0 - bH3 SnOs +2e+-S o+4,0tgH++ 40 Ht6e- 35n+ 3SnOs+be2N0+ -2 NO - + +Fe0s Te+L 6+b) +420 M2SOuM, +2+2502 - Stu + 4HtSOSOc+2HO2 +2e+-+4 0TeO2- H2OTe 36Ht1 +4e - ++3(SO2+4HzO+TeO +++6H-+e- 5,0(SOn Tee+ ++(NaOs INeCt(FeCts +2FeCleNazSnDsc) - +EQUIUBROAtAB =At① B B-0,8001= ABL=+ +-18,40.10-2 mol 0,105mof? Disso +7,30.10 X Xeq 7= +- AI0,105 7- Lw16-2m 37,30.10-=*EB]; e0,10 m== L3HzN2② i 1L2NMs v+ =600,0K 3Hz=>2NH3Tmol N20,783 = +L -Peg i75,48 -atm 38,57=? 2x 3xkc + X +-=PNHs+PNctPrzPeg 3738,57-27 Xeq= eq ea=0,783.0,0821.660,0Pres Peqatm 38,5738,37 3xx2x == += +-=1 4X38,57 -2x =+=kp=M(Pr)_ (18,46.3) 38,57 2x75,48(18 e = +NO=2,384. 19,93539,91 =2x x-=[38,57-(18,46.2)]2 1(RT)" *- 2,384.(0,0822.600)kp(RT)1p Dcx, - = == =55,381,65 ·Se eseiPr 1,65 ->=Pa 18,46=55,38PH, = Ceeg③ O p 3,0+ = 18= m 10.15 EI =**?mol⑩0ed risorsaProcessatePoi = 1 eper 1 aenCf

LOCK 4pCO 3,0=+ = =i 1806,2 -1231,5 XxA +-- 1231,Proeq 15-821821 410x1806,2x1231,5 Xep = =- - 985P(1, 1806,2-821 ==ea(exCO(g) (OCt,(8)- kp 3,0+ g) =1,00.10kmot- moe15 T22I =Pot=?mot10,0ed⑭ -N22NM3 M2 V3 OL400,=+i 20,0%.--0,100 ==?Rc3xX0,100-2xeq 0,100hNM3i=27 =0. 0,01=-10 x%. =->0,100[NM3] 0,082.0,010,100f 3= (0,01-4,22.1= - e[NcIf n.0.07 ==[H2]f 3.0,01 0,03= =-⑤ NONOz = SO3SOL I+ I2,0 alCON.4,0 5,04,0 eqz291 =101,0 ++ 6,040-x4,0-x x6,0+x +eqz x)(3,0 x)(6 +-28 +0,625k = (u x)= = x)(h --x23x6x18 ++ +0,625 16= x24x 4x +++⑲ borkpNcOusg) 0,66- atm0,6512 NOL (9) = =Prot Pa104 atm5,07 0,500· == PNor=?PronePo -E NcOn=?%.Up = I tutteche lesufficiente= specie sianoPat gas.PNiOnPNOz= + atm: 07.10"Pa5,1 atm:101325 Pa x=1045,07.S PotPNOn PNoz atm 0,500x -1323- === ses-xup-to* PNzOn065 0,331x= = =PNo2 atm0,1690,500-0,331 ==Proz wo: as-beigeNibu 20,311%. == ro, PNozPront⑰ E2502 deV0 2803 2,00+ = batPi mol 1,25totmol1 -2 =P kp=?bar0,25eq == xPsis: Plot. bar= barbar.0,25 1,25

0,27=+Xnathot= 2 2xso, 1User non x 3 2,73x2x-4ses =-=- +- -= =↓ t&Se2 1-27 nOn=2-7= 0,46=Psoz barbar.Pot pene0,2111,25 == 2,73 -bar. zanbar=Prot.Da 0,792in == 1,77=⑫ dunCO2COCCz 15,0CO v =+- > T1073k= barProt=mol0,10 5,0eq L bar2,17kp = Pco-Pal=?Pc1 atm Prock0,587= -bar=0,594 bar atm1 0,987=E Prot=Paz Pasalz Ptot-Pao-PolzPast Pocke E+ =ee eUp=Pepe,Kp * 217-= X5,0-0,594 -barPc7 3,45==Procer bar0,956=URCe⑬ +2020. 240I+ C.S.dundenin 673k2 T = PaProtdan1,05 1,01.10eq = dUp=?Detekmol barVm =1,0122,7 R 8,314==nace-suy= mol-PrC; berPa 0,6161861618 =0046 == mol barPa123235 1,23235=-Poni=nor= 0,0892mol 64657 0,64657 barPa0,0468 PCence = =->=UNC 2M20+IC2+02 =i --1,2320,616 - 0,6472x-4eq0,616 1,232 2x7 ->x =-POL 0,646)1,01-1,900-0,647/2 Price0,9091,232 == +=⑳ IN denE H2TNH3 1,5V+ = K448T = =?i Up1,29 15,4%NM3%. =2.1 mok[NH3] 0,0587== 1,5L17,0 319/moe in-hposoto 3= nol]TH2 0,0136=(900452)*.0,0136)kc =0,00215-= 0,00904)(0,0587 - (0,0821-448)" 0,079kp 0,00215. -=⑥ Mc

dun(0,g)MeOgsE 1,3Cit V =19)+ atm1bar 0,9871273k- TnI =n = barX PH,0 atm25 24,7Xxeq n = =- molnot=24Skp=?atm)PaPM_(71,25 atmkp = =205atm7zu,PM,Oha=-atm5s e0,354=-not.-x)(nntot 2x2x 0,3545 +2,4+ -> == 1,022.0,082.1273me RTPx Px·4,0 =1,022 hn4c -x = === == 1,5atm71,2=⑦ 3M20CuSOn CuSOv'3H10,) H2O2-· (g)(s) =?5,00g mdu5,022° 2pH02 mo2,0n.n34kp - =0,95.1 n- ,0 == = 3.18,016 0,036g-162,01himmm -== =⑧ NMs diss=20%2 H2N2 3 1LV7. =+ -= V=?i %diss=X 40% -34Ix-24ed ↓ N2 3H2NHs =20,1X +0,2.X-24 x= = X[NH3] [H2][Nc]0,8x 0,10x 0,3 x= == Y y 34x -=10,3x)3.0,10x ↳a,2.10-3 2y 0,2x0,4x +Ka ==== (0,8x)2 = =[NMs] [N2] VV12x[Nc]↳ -=4.2.10-3 4,2-10-3Ki == - 5,35L-3= =NeOy I⑨ 2N02 22°T =i atm 20%1 1 =ea 1 2xx +- =? Pi atm3,0->↓ ==0,20* =Pro,PN, de0,81-0,20 0,40==NcOn 2NOz-3atmi p=02x⑳a 3-7 4x2-0,20x 0,360,6 0 x=- =%= ·100 12%= 6⑩ H20 -ka 1,3.102502 2H2S2 = I 302 =+I 2628 II 2,6-2x 2x 372,8 2xeq - 3(2x)2.(3x)Rc = 2~ 2,6 Ex)(2,8 2x)- -2Fe(s)⑰

3M,0(g) Fe,03(s) 3H(g)= 800C+ T+ =Kn45 0,35100 gg =V 10.0L=Xz=?mot2,5h4,0 =[H,0] XMc0=?mol0,25=x e- 35rc =0,35= - -3x 0,592[H0Jeq 1,1776x0,63x 0,148mol 1,50,15 X,n -==-= = 0,0344,f76 0,1483 xx[4] = =mol0,1 1,0 tro 0,th ==ea = 2NHsNz 3H2⑫ +deI moe10,0 mot mot 4,04,02,0= ed. 1?I moe= 5,0 edz[NMs]i[N2]1 [M2]mol/ mof, mok0,40.2 0, u== == 2NH3n 3HzNc e12.5 += 4,02,0 4,02,0 4,03I 4,0 27x- +b un mol=3,0Re 12,5 mol0,5=*== e 12,3-63 1,07.V212,5-> 12,5= =. 3,4v =pH① Ka=?:ion=?pH=motHA0,345 1,67 M3OtM20MA = -A+ +nos 1,67-18ed%ion= 6,19%-no == 0,3450,345-10-167[HA]eq mol/0,324== (67)[M30Yea[ATea 1,4.153Ka =THATea= 0,324 -seeneBHt+OHEB H2O② 0,75y.+ -=1,50.10-2 =?mol pHpOH=?3,75.10[B] 1,50.102.0,25 mot=eq = pOH=-log[or]TOHTeq mol,1,50.12.0,75 1,13.10 3,95== =100 pH 1 10,05pOk- == pHfn=?③ NaOHCH3COOM CHsCeONa+H20-+ 3am -a 17,3 ka 1,80.1020,0 =↓ mol0,3153,51.10-3n kwaon== d CH,COOCH COOH H2OOH-->+ + -5,51.103 5,51.10-33 -imol-5,31.10n = 5,51.10-3f --3,51.10-3[CH COIf=

mol=0,170,0375 COO- OHMosCHICOPHCHE I +i -0,147 -10-14 3,56-10-10Kb f= -0,1471,80.10- XXx= 5 -x2 x)5,56.10-70(0,0747 x2Klo= -> - =-0,0147 -x [04]/8,9.10-68,17-10-1111 1 18x24- - 5,56.10-5,36.108,77.10 x = =+ -- = [0H-))log7 8,95pH 10 -& ==④ pHM,0OH - AHA + 8,33+ -> ==?aau ka17,825,0↓ mol0,096 5,67 2,14.10-3[OHJE --pH 101 5,67pOH↓nown = === 103 mol1,77n =17-10-3moe =[1] TAJeq mof,moll 0,03992,14.10=0,0400 0,0400= -= 0.0428 L emalore(i14.16 =-10Kb 8,71.10-5Ka=1,15.10= - -0,0399NH,S MeO OHENM3 12t V+ = 515,40.10-2- kb- -1,80.10- - =5,40.10-2-x ? pHXeq X =2,80.10-5_no