ГДЗ-Химия-9-класс-Дидактические-материалы-Радецкий-2000 (991555), страница 5

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KOH + HNO3 → KNO3 + H2Om( KOH )20=n(KOH) == 0,357 fhevM ( KOH ) 56m( HNO3 )25=n(HNO3) == 0,397 fhevM ( HNO3 ) 63LdHNO3 gZoh^blky\ ba[uld_lhjZkq_l\_^_fihKOH.20 ]o]KOH + HNO3 → KNO3 + H2O56 ]101 ]5420x=56 10120 ⋅ 101x== 36,1 ]56Hl\_lm(KNO3) = 36,1 ]<ZjbZglZ2Na + S → Na2S[ Ca + Cl2 → CaCl2\ 2Li + 2H2O → 2LiOH + H2↑2. Ca(OH)2 + CO2 → CaCO3↓ + H2O2KOH + CO2 → K2CO3 + H2OBa(OH)2 + CO2 → BaCO3↓ + H2O;Zjblh\Zy\h^Zlh`_[m^_lfmlg_lvgZ\ha^mo_tR3.

CaCO3 → CaO + CO2ω(CaCO3) = 100 – 8% = 92%m(CaCO3) = 1000 ⋅ 0,92 = 920 ]920 ]o]tRCaCO3 → CaO + CO2100 ]56 ]920x=100 56920 ⋅ 56x== 515,2 ]100Hl\_lm(CaO) = 515,2 ]<ZjbZglZ[\1. Na → NaOH → NaHCO3 → Na2SO4Z 2Na + 2H2O → 2NaOH + H2[ NaOH + CO2 → NaHCO3\ 2NaHCO3 + H2SO4 → Na2SO4 + 2H2O + 2CO2↑2. @_kldhklv\h^u²kh\hdmighklvk\hckl\h[mkeh\e_gguogZebqb_f\ \h^_bhgh\Mg2+ b Ca2+; kmffZjgh_kh^_j`Zgb_wlbobhgh\gZau\Z_lkyh[s_c`_kldhklvxJZaebqZxldZj[hgZlgmxbg_dZj[hgZlgmx`_kldhklvtRCa(HCO3)2 → CaCO3↓ + H2O + CO2Ca(HCO3)2 + Ca(OH)2 → 2CaCO3↓ + 2H2O11,5 ]78 ]3. 2Na + 2H 2 O → 2NaOH + H2↑P 1D =n(Na) == 0,5 fhev0 1D 55n(H2O) =P + 2 == 4,33 fhev0 + 2 LdG2H gZoh^blky\ ba[uld_jZkq_l\_^_fihNa.11,5 ]o]2Na + 2H2O → 2NaOH + H2↑2⋅23 ]2⋅40 ][= ⋅ ⋅ ⋅ x== 20 ] ⋅ Hl\_lm(NaOH) = 20 ]<ZjbZgl1.

NaOH + HCl → NaCl + H2ONaOH + CO2 → NaHCO32NaOH + H2SO4 → Na2SO4 + 2H2O2. IjbgZ]j_\ZgbbfjZfhjjZaeZ]Z_lky:CaCO3 → CaO + CO2CaO + H2O → Ca(OH)23.o]3e2Na + 2H2O → 2NaOH + H 2 ↑2⋅23 ]22,4 e[= ⋅ ⋅ ⋅ = ]bebηG2) =x=Hl\_lηG2) = 89,3%.JZ[hlZ:exfbgbcb _]hkh_^bg_gby<ZjbZgl1.

:O&O2 → 2AlCl32Al + 3H2SO4 → Al2(SO4)3 + 3H22Al + Cr2O3 → Al2O3 + 2Cr2Al + 3CuSO4 → Al2(SO4)3 + 3CuZ2. Al(OH)3 → AlCl3Al(OH)3ZAl(OH)3 + 3HCl → AlCl3 + 3H2O[ AlCl3 + 3NaOH → Al(OH)3 + 3NaClAl2O356t°Al2O3 + 3H2O\ 2Al(OH)33. :exfbgbc²k_j_[jbklh[_eucf_lZeee_]dbchq_gvfy]dbc<ZjbZgl2Al1. Fe2 O3 +→Al2O3+2Fehdbkebl_ev\hkklZgh\bl_evAl2(SO4)3 + 3BaCl2 → 2AlCl3 + 3BaSO4↓Al(OH)3 + NaOH → Na[Al(OH)4]4Al+ 3O 2 →2Al2O3\hkklZgh\bl_evhdbkebl_ev2. Al(NO3)3 → Al(OH)3Al(NO3)3 + 3NaOH → Al(OH)3↓ + 3NaNO3Al2O3 → Al2(SO4)3Al2O3 + 3H2SO4 → Al2(SO4)3 + 3H2OAl → AlCl32Al + 3Cl2 → 2AlCl33. :exfbgbcijbf_gy_lky\ kZfhe_lhkljh_gbb^eyba]hlh\e_gbyijh\h^h\\ f_lZeemj]bbijbwlhfbkihevamxlky_]he_]dhklv\ukhdZywe_dljhijh\h^ghklvb \hkklZgh\bl_evgZykihkh[ghklv<ZjbZgl1.

2Al + 3Cl2 → 2AlCl32Al + 6HCl → 2AlCl3 + 3H2Al + FeCl3 → AlCl3 + FeAl2(SO4)3 + 3BaCl2 → 2AlCl3 + 3BaSO4↓Z[\2. Al → Al2O3 → Al2(SO4)3 → Al(OH)3Z4Al + 3O2 → 2Al2O3[ Al2O3 + 3H2SO4 → Al2(SO4)3 + 3H2O\ Al2(SO4)3 + 6NaOH → 2Al(OH)3↓ + 3Na2SO43. :exfbgbch[jZam_lijhqgmxhdkb^gmxie_gdmgZk\h_cih\_joghklbihwlhfmhgg_hdbkey_lky^Ze__<ZjbZgl1. 2Al + 3Br2 → 2AlBr32Al + 3S → Al2S32Al + 6H2O → 2Al(OH)3 + 3H2↑2Al + 2NaOH + 6H2O → 2Na[Al(OH)4] + 3H2↑2.

2Al + 6HCl → 2AlCl3 + 3H2↑AlCl3 + 3KOH → Al(OH)3↓ D&O3HCl + Al(OH)3 → AlCl3 + 3H2O3. GZih\_joghklbjZkdZe_ggh]hZexfbgbyf]gh\_gghh[jZam_lkyhdkb^gZyie_gdZdhlhjZyaZljm^gy_l^_nhjfZpbx57JZ[hlZ Blh]h\Zyihl_f_V<ZjbZgl1. Ca(OH)2 + CO2 → CaCO3 + H2OCa(OH)2 + 2HCl → CaCl2 + 2H2OCa(OH)2 + ZnCl2 → Zn(OH)2↓ + CaCl2( +)2. A 2Cl– – 2e– → Cl20( −)K Fe2+ + 2e– → Fe02H2O + 2e– → H2 + 2OH–we a−FeCl2 + H2O → Cl2 + Fe + H2 + Fe(OH)21)3. Al 4)Al2O32)Al(NO3)33)Al(OH)35)Al(OH)3AlCl31) 4Al + 3O2 → 2Al2O32) Al2O3 + 6HNO3 → 2Al(NO3)3 + 3H2O3) Al(NO3)3 + 3NaOH → Al(OH)3 + 3NaNO34) 2Al + 6H2O → 2Al(OH)3 + 3H2↑5) Al(OH)3 + 3HCl → AlCl3 + 3H2O4.

:exfbgbc²kZfucjZkijhkljZg_ggucf_lZeeb hlghkbl_evghobfbq_kdbbg_jl_gaZkq_lh[jZam_fhchdkb^ghcie_gdb:exfhl_jfby²kihkh[\hkklZgh\e_gbyf_lZeeh\babohdkb^h\ijbf_gyxl\ f_lZeemj]bb5.140 ]o]CaO + H2O → Ca(OH)256 ]74 ][= ⋅x== ]η(Ca(OH)2) == 0,984 bebHl\_lη(Ca(OH)2) = 98,4%.<ZjbZgl1. Al2O3 y\ey_lkyZfnhl_jgufhdkb^hfAl2O3 + 2NaOH → 2NaAlO2 + H2OAl2O3 + 6HCl → 2AlCl3 + 3H2O( +)2. A 2Cl– – 2e– → Cl2058( −)K Na+ + 1e– → Na0we a−2NaClj\ → 2Na + Cl25)2)1)3)4)3. Ca → CaCl2→ CaO → Ca(OH)2 →CaCO →Ca(HCO3)2 1) 2Ca + O2 → 2CaO2) CaO + H2O → Ca(OH)23) Ca(OH)2 + CO2 → CaCO3↓ + H2O4) CaCO3 + H2O + CO2 → Ca(HCO3)25) Ca(HCO3)2 + 2HCl → CaCl2 + 2H2O + 2CO2↑4.

DZevpbc²hq_gvfy]dbcb j_Zdpbhgghkihkh[gucf_lZee5. Ca(OH)2 + Na2CO3 → CaCO3↓ + 2NaOHω(Na2CO3) = 100% – 5% = 95%m(Na2CO3) = 80 ⋅ 0,95 = 76 ]76 ]o]106 ]100 ]Ca(OH)2 + Na 2 CO3 → CaCO3 ↓ + 2NaOH[= ⋅x== 71,7 ]Hl\_lm(CaCO3) = 71,7 ]<ZjbZgl1.

Al(OH)3 + NaOH → Na[Al(OH)4]Al(OH)3 + 3HCl → AlCl3 + 3H2OtR2Al(OH)3 → Al2O3 + 3H2O( +)2. A 2I– – 2e– → I20( −)K K+ + 1e– → Kwe a−2KIj\ → 2K + I25)2)4)1)3. CaCl2 → CaCO3 →Ca(HCO3)2 3)→CaCO3 →Ca(NO3)2 →CaCO31) CaCl2 + Na2CO3 → CaCO3 + 2NaCl2) CaCO3 + H2O + CO2 → Ca(HCO3)2tR3) Ca(HCO3)2 → CaCO3↓ + CO2 + H2O4) CaCO3 + 2HNO3 → Ca(NO3)2 + H2O + CO25) Ca(NO3)2 + Na2CO3 → CaCO3 + 2NaNO34. @_kldhklvh[mkeh\e_gZkh^_j`Zgb_f\ \h^_bhgh\Mg2+ b &D2+.tRCa(HCO3)2 → CaCO3 + H2O + CO2↑Ijhbkoh^blh[jZah\Zgb_gZdbibb aZkhj_gb_dhleZ595. CaCO3 → CaO + CO2↑ω(CaCO3) = 100% – 20% = 80%m(CaCO3) = 500 ⋅ 0,8 = 400 ]o]400 ]CaCO3 → CaO + CO2↑56 ]100 ][= ⋅ x== 224 ]Hl\_l]<ZjbZgl1.

KZHy\ey_lkyhkgh\gufhdkb^hfZCaO + CO2 → CaCO3[ CaO + H2O → Ca(OH)2\ CaO + 2HCl → CaCl2 + H2O( +)2. A 2Br– – 2e– → Br20( −)K Cu2+ + 2e– → Cu0we a−CuBr2 jj → Cu + Br23.1Al2O32Al2(SO4)33Al4AlCl35Al(OH)31) 4Al + 3O2 → 2Al2O32) Al2O3 + 3H2SO4 → Al2(SO4)3 + 3H2O3) Al2O3 + 6HCl → 2AlCl3 + 3H2O4) 2Al + 6HCl → 2AlCl3 + 3H2O5) AlCl3 + 3NaOH → Al(OH)3 + 3NaCl4. < ijbjh^_Zexfbgbc\klj_qZ_lky\ \b^_Al2O3 \ khklZ\_]ebgu[hdkblh\g_n_ebgZb qbklh]hdhjmg^Z3,15 ]o]2⋅63 ]164 ]5. Ca(OH)2 + 2HNO3 → Ca(NO3 ) 2 + 2H2O[= ⋅ ⋅x== 4,1 ⋅ 60mij(HNO3) = η(HNO3) ⋅ ml_hj = 4,1 ⋅ ]Hl\_lm(HNO3)2 = 4,02 ]61L_fZVI@_e_ahF_lZeemj]byJZ[hlZ@_e_ahb _]hkh_^bg_gby<ZjbZgl1. 4Fe + 3O2 → 2Fe2O3 hdkb^`_e_aZIII)2Fe + 3Cl2 → 2FeCl3 oehjb^`_e_aZIII)tRFe + S → FeS kmevnb^`_e_aZ(II)2Fe + 3Br2 → 2FeBr3 [jhfb^`_e_aZ(III)2.

+26Fe 1s22s22p63s23p63d64s2tR < kh_^bg_gbyo`_e_ahijhy\ey_lkl_i_gbhdbke_gby3. 3Fe2+ + 2K3[Fe(CN)6] → Fe3 [Fe(CN)6 ]2 ↓ + 6K+l_fghkbgbc4Fe3+ + 3K4[Fe(CN)6] → Fe 4 [Fe(CN)6 ]3 ↓ + 12K+32 ]28 ][bjxah\uc4. CuSO 4 + Fe → FeSO4 + Cun(Fe) == 0,5 fhevn(CuSO4) == 0,2 fhevLdCuSO4 gZoh^blky\ g_^hklZld_jZkq_l\_^_fihg_fm32 ]o]CuSO 4 + Fe → FeSO4 + Cu160 ][= ⋅ x== 12,8 ]64 ]Hl\_lm(Cu) = 12,8 ]<ZjbZgl1. Fe(NO3)3 → Fe(OH)3 → Fe2O3 → Fe → FeCl21) Fe(NO3)3 + 3NaOH → Fe(OH)3↓ + 3NaNO31)2)3)4)tR2) 2Fe(OH)3 → Fe2O3 + 3H2O3) Fe2O3 + 3H2 → 2Fe + 3H2O624) Fe + 2HCl → FeCl2 + H2↑2. Kh_^bg_gby`_e_aZkhkl_i_gvxhdbke_gbyijhy\eyxl\hkklZgh\bl_evgu_k\hckl\Zijbwlhfhdbkeyykv^hkl_i_gbhdbke_gbyldboZlhfijbwlhfi_j_oh^bl\ [he__mklhcqb\mxnhjfm< g_dhlhjuokemqZyowlbkh_^bg_gbyfh]mlijhy\eylvhdbkebl_evgu_k\hckl\Zdh]^Zbagbo\ul_kgy_lkyf_lZeebq_kdh_`_e_ah3.

Fe + 2HCl → FeCl2 + H2↑FeCl2 + 2NaOH → Fe(OH)2 + 2NaCl4Fe(OH)2 + 2H2O + O2 → 4Fe(OH)3112 ]oe4. Fe + H2SO4 → FeSO4 + H 256 ][= ⋅ x== 44,8 e22,4 eVijG2) = ηG2) ⋅ Vl_hj = 0,98 ⋅ eHl\_lV(H2) = 43,9 e<ZjbZglZFe + H2SO4 → FeSO4 + H2[ Fe + CuSO4 → FeSO4 + Cu\ Fe + 2HBr → FeBr2 + H2] Fe + Pb(NO3)2 → Fe(NO3)2 + Pb2. )H2b )H2O3 — hkgh\gu_hdkb^u)H2+2 b )H2+3 — hkgh\gu_]b^jhdkb^u3. Ij_^eh`_ggu_\_s_kl\Zfh`ghjZkihagZlvijbihfhsbjZkl\hjh\1D2SO4 b NaOH.FeCl2 + 2NaOH → Fe(OH) 2 ↓ + 2NaCla_e_gucBaCl2 + Na2SO4 → BaSO 4 ↓ + 2NaCl[_euc< hklZ\r_ckyijh[bjd_[m^_lNaCl.4.

2HCl + FeCO3 → FeCl2 + H2O + CO2m(FeCO3) = 0,8 ⋅ 50 = 40 ]40 ]oe116 ]22,4 e2HCl + FeCO3 → FeCl2 + H2O + CO 2[= 63x= ⋅ = 7,72 eHl\_lV(CO2)=7e<ZjbZgl1. Fe → FeSO4 → Fe(OH)2 → Fe(OH)3 → FeCl31) Fe + H2SO4 → FeSO4 + H22) FeSO4 + 2NaOH → Fe(OH)2↓ + Na2SO43) 4Fe(OH)2 + 2H2O + O2 → 4Fe(OH)34) Fe(OH)3 + 3HCl → FeCl3 + 3H2O2. <h^gu_jZkl\hjukhe_c`_e_aZ(III) bf_xldbkemxkj_^mldkhebh[jZah\ZgukeZ[ufhkgh\Zgb_fb kbevghcdbkehlhcb \\h^_]b^jhebamxlkyZFeSO4 + BaCl2 → FeCl2 + BaSO4↓[ FeSO4 + Ba(NO3)2 → BaSO4↓ + Fe(NO3)24.1)7]2)3)4)18 ]2Fe + 3Cl2 → 2FeCl3P )H== 0,125 fhev0 )H P&O n(Cl2) === 0,253 fhev0 &O n(Fe) =LdCl2 gZoh^blky\ ba[uld_jZkq_l\_^_fihFe.7]o]2Fe + 3Cl2 → 2FeCl32⋅56 ]2⋅162,5 ][= ⋅ ⋅ ⋅ ⋅ x== 20,31 ] ⋅ Hl\_lm(FeCl3) = 20,31 ]JZ[hlZKihkh[uijhfure_ggh]hihemq_gbyf_lZeeh\<ZjbZglZ`_e_agZyjm^Z)H2O3)[1) ihemq_gb_\hkklZgh\bl_ey\2C + O2 → 2COFe2O3 + 3CO → 2Fe + 3CO22) \hkklZgh\e_gb_tR3) CaCO3 → CaO + CO23) reZdhh[jZah\Zgb_64CaO + SiO2 → CaSiO33CaO + P2O5 → Ca3(PO4)2MnO + SiO2 → MnSiO32.o]0,1 ]C + O2 → CO 212 ]44 ][ = ⋅ x== 0,0273 ]P& =ωKbebPklHl\_lωK <ZjbZglZqm]mg[\2Fe + O2 → 2FeOSi + 2FeO → SiO2 + 2FeMn + FeO → MnO + Fe2P + 5FeO → 5Fe + P2O5C + FeO → Fe + CO2.

2Al2O3 → 4Al + 3O2ω(Al2O3) = 100% – 0,4% = 99,6%m(Al2O3) =0,996 ⋅ 50 l l49,8 ]o]2Al 2 O3 → 4Al + 3O22⋅102 ]4⋅27 ][= ⋅ ⋅ ⋅ ⋅ x== 26,36 l ⋅Hl\_lm(Al) = 26,36 l<ZjbZglZAl2O3 b Na3[AlF6] ([hdkblu[→ 2NaAlO2 + CO2\Al2O3 + Na2CO3 2NaAlO2 + CO2 + 3H2O → 2Al(OH)3 + Na2CO32Al(OH)3 → Al2O3 + 3H2OtR65we a−2Al2O3 → 4Al + 3O2:exfbgbcijbf_gy_lky\ kZfhe_lhkljh_gbbf_lZeemj]bbb ^eyba]hlh\e_gbyijh\h^h\2.o]1,96 eFe + H2SO4 → FeSO4 + H 256 ]22,4 e[= ⋅x== 4,9 ]ω(Fe) == 0,98 bebHl\_lω(Fe) = 98%.<ZjbZglZqm]mg[\2Fe + O2 → 2FeOFeO + C → Fe + CO2FeO + Si → 2Fe + SiO25FeO + 2P → 5Fe + P2O5FeO + Mn → MnO + Fe2. ] K²] Fe3C) — x% ⋅x== 54%Hl\_lω(Fe3C) = 54%.66AZ^Zqbk ijhba\h^kl\_ggufb f_`ij_^f_lgufkh^_j`Zgb_fG_f_lZeeuR1.

SiO2 + 2C → Si + 2COm(C) = 80 ⋅ 0,92 ] d]to d]73,6 d]tRSiO 2 + 2C → Si + 2CO60 d]2⋅12 d][= ⋅ ⋅ x== 184 d] ⋅ Hl\_lm(SiO2) = 184 d]o d]Rm f3248 d]t2. 2Ca3(PO4)2 + 10C + 6SiO2 → 6CaSiO3 + 10CO2 + P4120 d]224 f124 d]x248=120 124120 ⋅ 248= 240 d]Ko124m=224 124224 ⋅ 248mf3 KH124mij (C) 240ml_hjK=d]ω(C)0,9Hl\_lm(C) = 266,7 d]V(CO) = 448 f3.o d]3. P4124 d]400 d]+ 5O2 + 6H2O → 4H 3 PO 44⋅98 d]x400=124 4 ⋅ 98124 ⋅ 400x== 126,5 d]4 ⋅ 98ml_hjJmij (P)η(H3 PO4 )=126,5d]0,867Hl\_lm(P) = 158,2 d]120 d]tRo d]4.

Ca 3 (PO 4 ) 2 + 3H2SO4 → 3CaSO4 + 2H 3 PO 4310 d]2⋅98 d]m(Ca3(PO4)2) = 0,6 ⋅ 200 = 120 d]120x=310 2 ⋅ 98120 ⋅ 2 ⋅ 98od]310mij(H3PO4) = ml_hj ⋅ η(H3PO4) = 0,8 ⋅ 75,87 = 60,7 d]Hl\_lm(H3PO4) = 60,7 d]5. SO2 + NO2 + H2O = H2SO4 + NOmjjZ = ρ ⋅ V = 1,66 ⋅ 100 = 166 d]m(H2SO4) = ω ⋅ mjjZ = 166 ⋅ d]o d]m d]122,84 d]SO 2 + NO2 + H 2 O = H 2SO 4 + NO64 d]18 d]98 d]x122,84=649864 ⋅ 122,84od](SO2)98m =189818 ⋅ 122,84= 22,6 d]G2Hm98Hl\_lm(SO2) = 80,22 d]m(H2O) = 22,6 d]6. ∆ωJ ± 2,13%ωG2Hbeb2,51%Hl\_lωG2H 7.

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