ГДЗ-Химия-9-класс-Дидактические-материалы-Радецкий-2000 (991555), страница 3
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P → Ca3P2 → PH3 → P2O5Z[\Z3&D → Ca3P2[&D3P2+6H2O → 3Ca(OH)2+2PH3\3+3+4O2 → P2O5+3H2O<ZjbZglZ3&D → Ca3P2t0[&D3(PO4)2+3H2SO4 → 3CaSO4+2H3PO4\1D2++3PO4 → Na3PO4+3H2OOH-+H+ → H2OIjbk]hjZgbb[_eh]hb djZkgh]hnhknhjZ\ h^bgZdh\uomkeh\byoh[jZamxlkyh^bgZdh\u_\_s_kl\Z<h^mfh`ghhij_^_eblvijbihfhsbeZdfmkZ\h^Zbf__lg_cljZevgmxkj_^mb eZdfmk\ g_chklZg_lkynbhe_lh\uf\dbkehlZohgihdjZkg__l+NO3\ hlebqb_hl+3PO4, jZkl\hjy_lf_^vCu+HNO3 → Cu(NO3)2 ↓ +2NO2+2H2O`_elucCu+H3PO4 →/29<ZjbZglZ36 → P2S3[32O5+3H2O → 2H3PO4\+3PO4+3Na2CO3 → 2Na3PO4+3CO2+3H2O2H++CO − → CO2+H2OFhe_dmeZnhknhjZ\ iZjZokhklhblbao Zlhfh\34Z fhe_dmeZZahlZbao 12)Z[\3. P → P2O5 → H3PO4 → K3PO4Z322 → 2P2O5[32O5+3H2O → 2H3PO4\.2++3PO4 → K3PO4+3H2OJZ[hlZJZkq_luihhij_^_e_gbxfZkkh\hcbebh[t_fghc^heb\uoh^Zijh^mdlZj_Zdpbbhll_hj_lbq_kdb\hafh`gh]hb h[jZlgu_aZ^Zqb252 ]o]63 ]85 ]1.
NaOH+ HNO3 → NaNO3 +H2O252 o ⋅ o]=63 85mij(NaNO3)= § (NaNO3) ⋅ ml_hj(NaNO3)=0,9 ⋅ ]Hl\_lmij(NaNO3 ]20 ]oe2⋅53,5 ]2⋅22,4 e2. 2NH 4 Cl +Ca(OH)2 → CaCl2+ 2NH 3 +2H2O20o ⋅ ⋅ oe= ⋅ 2 ⋅ 53,5 2 ⋅ 22, 4Vij(NH3)= § (NH3) ⋅ Vl_hj(NH3)=8,37 ⋅ eHl\_l91+3 e280 ]o]3. CaO +H2O → Ca(OH)256 ]280 o=56 74o§ (Ca(OH)2))=74 ] ⋅ ]m ij (Ca(OH) )m l_hj (Ca(OH) )Hl\_l § (Ca(OH)2))=96,76%20 ]=358beb 37020 ]4. 2NH 4 Cl + CaO → 2NH3+CaCl2+H2O30n(NH4Cl)=n(CaO)=m(CaO)20=fhevM(NH 4Cl) 53,5m(CaO) 20=fhevM(CaO) 56Ld&D2gZoh^blky\ ba[uld_jZkq_l\_^_fih1+4Cl.20 ]o]2⋅53,5 ]2⋅17 ]2NH 4 Cl +CaO → 2NH 3 +CaCl+H2O20o ⋅ ⋅=o] ⋅ 2 ⋅ 53,5 2 ⋅17mij(NH3)= § (NH3) ⋅ ml_hj(NH3)=0,98 ⋅ ]Hl\_lmij(NH3 ]6,4 ]oetR5. NH 4 NO 2 → N 2 +2H2O64 ]22,4 e6, 4o ⋅ =oe64 22, 4Vij(N2)= § (N2) ⋅ Vl_hj=0,89 ⋅ eHl\_lVij(N2 e56 ]oe22,4 e2⋅22,4 e6.
N 2 +3H2 → 2NH 356o ⋅ ⋅ =oe22, 4 2 ⋅ 22, 4Vij(NH3)= § (NH3) ⋅ Vl_hj(NH3)=0,5 ⋅ e9m(NH3)=⋅ m(NH3)=⋅ ]9PHl\_lVij(NH3 em(NH3 ]107 ]oeot7. NH 4 Cl → NH 3 +HCl53,5 ]22,4 eo= § (NH3)=VijVl_hjo= ⋅ ebebHl\_l § (NH3)=84,8%31o d]200 d]8. P → H 3 PO 431 d]107o=53,5 22, 4mij(P)=98 d]m l_hjη(P)=o⋅ d]]Hl\_lP3 ]33 ]oe132 ]2⋅22,4 e9.
(NH 4 )2SO4 +Ca(OH)2 → CaSO4+ 2NH 3 +2H2O33o ⋅ ⋅ oe=132 2 ⋅ 22, 4Vij= § (NH3) ⋅ Vl_hj(NH3)=11,2 ⋅ eHl\_l91+3 eo]4e10. Cu+ 4HNO3dhgp → Cu(NO3)2+ 2NO 2 +2H2O64 ]o=64 44,8mij(Cu)=m l_hjη(P) ⋅ o]=2⋅22,4 e]Hl\_lm(&X ]30 ]oe2⋅53,5 ]2⋅22,4 e11. 2NH 4 Cl +Ca(OH)2 → CaCl2+ 2NH 3 ↑ +2H2O33o ⋅ ⋅ oe= ⋅ 132 2 ⋅ 22, 4Vij(NH3)=Vl_hj ⋅ § (NH3)=12,561 ⋅ eHl\_l91+3 e20,2 ]o]2⋅101 ]2⋅63 ]12. 2KNO3 +H2SO4 → K2SO4+ 2HNO320, 2o ⋅ ⋅ o]= ⋅2 ⋅101 2 ⋅ 63mij(HNO3)=ml_hj ⋅ § (HNO3)=0,98 ⋅ ]Hl\_lm(HNO3 ]3240 ]oeot13. NH 4 NO3 → N 2 O +2H2O22,4 e80 ]40o ⋅ =oe80 22, 4Vij(N2O)=Vl_hj ⋅ § (N2O)=11,2 ⋅ eHl\_l9122 e80 ]o]310 ]2⋅98 ]14.
Ca 3 (PO 4 ) 2 +3H2SO4 → 3CaSO4+ 2H 3 PO 480o ⋅ ⋅ o]=310 2 ⋅ 98mij(H3PO4)= § (H3PO4) ⋅ ml_hj=0,96 ⋅ ]Hl\_lm(H3PO4 ]<ZjbZgl1.JZ[hlZBlh]h\Zyihl_f_,,,IjbagZdbkjZ\g_gbyZ:]j_]Zlgh_khklhygb_gm[Obfbq_kdb_k\hckl\Z]Za:ahl1) N2+3H2 → 2NH32) N2+O2 → 2NO3) N2+3Ca → Ca3N24) 2N2+3Si → Si3N4\J_ZdpbhggZykihkh[ghklbb^_ckl\b_gZhj]Zgbafq_eh\_dZObfbq_kdbfZehZdlb\_gmqZkl\m_l\ ijhp_kk_^uoZgbyNhknhj[_eucL\_j^h_\_s_kl\h1)2P+8H2O →→ 2H3PO4+5H22) 2P+5O2 → 2P2O53) 2P+3Ca → Ca3P24) 2P+5Cl2 → 2PCl5Obfbq_kdb\ukhdhZdlb\_gy^h\bl5)2. P → P2O5 → HPO3 → H3PO4 → Na3PO4 → Ag3PO41) 4P+5O2 → 2P2O52) P2O5+H2O → 2HPO33) HPO3+H2O → H3PO44) H3PO4+3NaOH → Na3PO4+3H2O5) Na3PO4+3AgNO3 → Ag3PO4 ↓ +3NaNO31)2)3)4)3337 ]o]3.
Ca(OH)2 +(NH4)2SO4 → 2NH 3 +CaSO4+2H2O74 ]37o=74 2 ⋅ 17§ (NH3)=2⋅17 ]m ijm l_hj ⋅ ⋅o]=bebHl\_l § (NH3)=88,24%<ZjbZgl1.IjbagZdbkjZ\g_gbyZKljh_gb_Zlhfh\:ahl1S22S22P3+7N+15P+7[<Z`g_crb_kl_i_gbhdbke_gby\KhklZ\b k\hckl\Zhdkb^h\Nhknhj[_euc1s22s22p63s23p3+1525GZb[he__jZkijhkljZg_gukh_^bg_gbykhkl_i_gyfbhdbke_gby±3, 0, +4, +5H[jZam_lhdkb^u12O,NO, N2O3, NO2, N2O5.GZb[he__mklhcqb\uN22±[_kp\_lguc]ZakijbylgufaZiZohfbkeZ^dh\Zluf\dmkhf12±[_kp\_lguc]Za[_a\dmkZb aZiZoZ122±[mjuc]Za285GZb[he__jZkijhkljZg_gukh_^bg_gbykhkl_i_gyfbhdbke_gby±H[jZam_lhdkb^u32O3 bP2O5.P2O3 ±[_eh_djbklZeebq_kdh_\_s_kl\hP2O5 ±[_eucihjhrhd2. N2 → NH3 → NO → NO2 → HNO3 → Ba(NO3)21)2)3)4)5)1) N2+3H2 → 2NH3Pt2) 4NH3+5O2 → 4NO+6H2O3) 2NO+O2 → 2NO24) 4NO2+2H2O+O2 → 4HNO35) 2HNO3+Ba(OH)2 → Ba(NO3)2+2H2O]o]3.
2NaNO3+H2SO4 → Na2SO4+2HNO32 ⋅ ]2 ⋅ ]o ⋅ ⋅ =o] ⋅ ⋅ ⋅ mij(HNO3)= § (HNO3) ⋅ ml_hj=12,6 ⋅ ]Hl\_lm(HNO3 ]Pt34<ZjbZgl1.IjbagZdbkjZ\g_gbyZKljh_gb_fhe_dmeu[Nbabq_kdb_k\hckl\Z\Obfbq_kdb_k\hckl\ZNH3PH3H→N←HH-P-H|H↑H=Za[_ap\_lZk j_adbfaZiZohf ;_kp\_lguc]Zak q_kghqgufe_]q_\ha^moZohjhrhaZiZohfly`_e__\ha^moZjZkl\hjbf\ \h^_fZehjZkl\hjbf\ \h^_hq_gvy^h\bl1) NH3+H2O → NH4OH1) 2PH3+4O2 → P2O5+3H2O2) 4NH3+3O2 → N2+6H2O2) PH3+HCl → PH4Cl3) NH3+HCl → NH4Cl2. P → CaP2 → PH3 → P2O5 → HPO3 → H3PO41) 2P+3Ca → Ca3P22) Ca3P2+6H2O → 3Ca(OH)2+2PH33) 2PH3+4O2 → P2O5+3H2O4) P2O5+H2O → 2HPO35) HPO3+H2O → H3PO41)2)3)4)30 ]oe2⋅53,5 ]2⋅22,4 e5)3. 2NH 4 Cl +Ca(OH)2 → CaCl2+ 2NH 3 +2H2O30o ⋅ ⋅ o]= ⋅ 2 ⋅ 53,5 2 ⋅ 22, 4Vij(NH3)= § (NH3) ⋅ Vl_hj=0,92 ⋅ eHl\_l91+3 e<ZjbZgl1.HNO3IjbagZdbkjZ\g_gbyZNbabq_kdb_ ;_kp\_lgZy^ufysZykygZk\hckl\Z\ha^mo_`b^dhklvtie=-420Ktdbi=830K[Obfbq_kdb_ 1) 8HNO3+3Cu → 3Cu(NO3)2+k\hckl\Z+2NO+4H2OH3PO4+Cu →/2) 10HNO3+4Zn → 4Zn(NO3)2++NH4NO3+3H2O3) HNO3+NaOH → NaNO3+H2OH3PO4;_eh_djbklZeebq_kdh_\_s_kl\htie=420K1) 2H3PO4+3Zn → Zn3(PO4)2+3H2 ↑2) H2PO4+3NaOH → Na3PO4+3H2O2)1)4)2.
Pb(NO3)2 →NO2 →NH4NO3 →(NH4)2SO4→NH3 →HNO3 3)5)tR1) 2Pb(NO3)2 → 2PbO+4NO2+O2352) 4NO2+2H2O+O2 → 4HNO33) HNO3+NH3 → NH4NO34) NH4NO3+NaOH → NH3+H2O+NaNO35) 2NH3+H2SO4 → (NH4)2SO462 ]o]3. 4P +5O2 → 2P2 O54⋅31 ]62o=4 ⋅ 31 2 ⋅142§ (P2O5)=m ijm l_hj2⋅142 ] ⋅ ⋅ ] ⋅ =beboHl\_l § (P2O5)=91,55%36L_fZ ,9Ih^]jmiiZm]e_jh^ZJZ[hlZM]e_jh^Hdkb^um]e_jh^Z<ZjbZgl1. K1s22s22p2 * K↑↓ ↑↓↑ ↑1s 2s 2p↑↓ ↑ ↑ ↑ ↑I_j_oh^ \hafh`_g ld bf__lky k\h[h^gZy hj[blZev ijb wlhfjZagbpZ\ wg_j]bbf_`^m2s- b p-hj[blZeyfbg_\bebdZZCuO+C → Cu+CO[&&22 → 2COt0\&D&23 → CaO+CO2 ↑3. 2CO+O2 → 2CO2¹ (CO)=100%-5%=95%V(CO)=V]ZaZ ⋅ ¹ (CO)=20 ⋅ eeoe2CO + O2 → 2CO22 ⋅ e eo ⋅ =oe ⋅ Hl\_l922 e<ZjbZglM]e_jh^\ ijbjh^_\klj_qZ_lky\ \b^_ZZefZaZ±hq_gvl\_j^h_ijhqgh_\_s_kl\hk \ukhdhctiedjbklZeebq_kdZyj_r_ldZ±ZlhfgZy[]jZnblZ±l\_j^h_\_s_kl\h`bjgh_gZhsmivk \ukhdhctiewe_dljhb l_iehijh\h^_gdjbklZeebq_kdZyj_r_ldZZlhfgZy\dZj[bgZ±l\_j^h_\_s_kl\hl\_j`_]jZnblZghfy]q_ZefZaZq_jgucf_edhdjbklZeebq_kdbcihjhrhdh[eZ^Z_lihemijh\h^gbdh\ufbk\hckl\ZfbDjbklZeebq_kdZyj_r_ldZZlhfgZyZ&22+Ca(OH)2 → CaCO3+H2O213[&+2 → CH4\&22 → CO23.
CaCO3+2HNO3 → Ka(NO3)2+H2O+CO2¹ (CaCO3)=100%-8%=92%m(CaCO3)=0,92 ⋅ ]tR3755,2 ]oe100 ]22,4 eCaCO3 +2HNO3 → Ca(NO3)2+H2O+ CO 255, 2o ⋅ oe=100 22, 4Hl\_l9&22 e<ZjbZglM]e_jh^\ kh_^bg_gbyoijhy\ey_lkl_i_gbhdbke_gby±+2, +4C-4H + , C − H + , C+2O-2, C+4O − ZCO+FeO → Fe+CO2[&$O → Al4C3\&22+CaO → CaCO33.
C+O2 → CO2¹ (C)=100%-6%=94%m(C)=0,94 ⋅ ]]oeC+O2 → CO2]e ⋅ o=oeHl\_l9&22 e<ZjbZgl:^khj[pby²ijhp_kkih]ehs_gbyih\_joghklvx\_s_kl\Z^jm]h]h\_s_kl\Z>_khj[pby²ijhp_kk\u^_e_gby\_s_kl\Zk _]hih\_joghklb:^khj[pbhggZykihkh[ghklv^j_\_kgh]hm]eyijbf_gy_lky\f_^bpbg_ijbhljZ\e_gbyo^eyih]ehs_gbyy^h\Z&222 → 2CO2[&6 → CS2\&D&23+2HCl → CaCl2+H2O+CO2tR3. CaCO3 → CaO+CO2¹ (CaCO3)=100%-10%=90%m(CaCO3)=0,9 ⋅ ]]oe450 ]oe100 ]22,4 eCaCO3 → CaO+ CO 2450o=100 22, 4o ⋅ e38Hl\_l9&22 eJZ[hlZM]hevgZydbkehlZb __kheb<ZjbZgl1. CaCO3 → CO2 → CaCO2 → Ca(HCO3)2 → CaCl21)2)3)4)tR1) CaCO3 → CaO+CO22) CO2+Ca(OH)2 → CaCO3+H2O3) CaCO3+H2O+CO2 → Ca(HCO3)24) Ca(HCO3)2+2HCl → CaCl2+2H2O+2CO2Kh^_j`bfh_ijh[bjhdfh`ghjZkihagZlvijbihfhsbjZkl\hjZAgNO3.AgNO3+KCl → AgCl ↓ +KNO3[_euc2AgNO3+K2CO3 → Ag 2 CO3k\_leh`_eluc↓ +2KNO33AgNO3+K3PO4 → Ag3 PO 4 ↓ +3KNO3`_eluc3.
CO2+H2O → H2CO3H[jZamxsZykydbkehlZhdjZrb\Z_leZdfmk\ djZkgucp\_lijbgZ]j_\ZgbbdbkehlZjZaeZ]Z_lkyihwlhfmeZdfmk\gh\vklZgh\blkynbhe_lh\ufH2CO3 → H2O+CO2 ↑<ZjbZgl1. CO → CO2 → KHCO3 → K2CO3 → CO21) 2CO+O2 → 2CO22) CO2+KOH → KHCO33) KHCO3+KOH → K2CO3+H2O4) K2CO3+2HCl → 2KCl+H2O+CO2IjbijhimkdZgbbm]e_dbkeh]h]ZaZq_j_aba\_kldh\mx\h^mkgZqZeZgZ[ex^Z_lky__ ihfmlg_gb_Z ihlhfjZkl\hj_gb_hkZ^dZCa(OH)2+CO2 → CaCO3 ↓ +H2OCaCO3+CO2+H2O → Ca(HCO3)2<h^gu_jZkl\hjudZj[hgZlh\dZebyb gZljbybf_xls_ehqgmxkj_^mldwlbkhebh[jZah\Zgukbevgufhkgh\Zgb_fb keZ[hcdbkehlhcb \\h^_]b^jhebamxlky<ZjbZgl1)2)3)4)1. Ca(HCO3)2 → CaCO3 → CO2 → Na2CO3 → H2CO31)2)3)4)R1) Ca(HCO3)2 → CaCO3+H2O+CO2t392) CaCO3+2HCl → CaCl2+H2O+CO23) CO2+2NaOH → Na2CO3+H2O4) Na2CO3+2HCl → 2NaCl+H2CO3tR2.
CaCO3 → CaO+CO2CO2+KOH → KHCO3tR2KHCO3 → K2CO3+H2O+CO2Khebm]hevghcdbkehlufh`ghhlebqblvijbihfhsbj_Zdpbbkkbevghcdbkehlhcijbwlhf[m^_l\u^_eylvkym]e_dbkeuc]Za<ZjbZgl1. C → CO2 → H2CO3 → Na2CO3 → NaNO31) C+O2 → CO22) CO2+H2O → H2CO33) H2CO3+2NaOH → Na2CO3+2H2O4) Na2CO3+2HNO3 → 2NaNO3+H2O+CO2 ↑ZBa(OH)2+Na2CO3 → BaCO3 ↓ +2NaOH[+2SO4+Na2CO3 → Na2SO4+H2O+CO2 ↑Ihemqblvkh^mlZdbfh[jZahfg_evayld&D&23 ±g_jZkl\hjbfZy\ \h^_khev1)2)3)4)JZ[hlZDj_fgbcb _]hkh_^bg_gby<ZjbZglJZ^bmkZlhfZdj_fgby[hevr_jZ^bmkZZlhfZm]e_jh^Z_]hwe_dljhgukeZ[__ijbly]b\Zxlkyd y^jml_hge_]q_bohl^Z_lihlhfm_]hg_f_lZeebq_kdb_k\hckl\Z\ujZ`_gukeZ[__Z[\2.
SiO2 → Si → SiO2 → Na2SiO3→ Si+CO2Z6L22+C [6L22 → SiO2\6L22+2NaOH → Na2SiO3+H2O3. Mg2Si+4HCl → 2MgCl2+SiH4¹ (Mg2Si)=100%-5%=95%m(Mg2Si)=0,95 ⋅ ]tR152 ]o]76 ]32 ]Mg 2Si +4HCl → 2MgCl2+ SiH 4152 o ⋅ o]=76 32Hl\_lm(SiH4 ]40<ZjbZgl<h^gu_jZkl\hjukbebdZlh\gZljbyb dZebybf_xls_ehqgmxkj_^mldwlbkhebh[jZah\Zgukbevgufhkgh\Zgb_fb keZ[hcdbkehlhcb \\h^_]b^jhebamxlkyZ[\2. NaOH → Na2SiO3 → H2SiO3 → SiO2Z1D2+6L22 → Na2SiO3+H2O[1D2SiO3+2HCl → 2NaCl+H2SiO3 ↓\+2SiO3 → H2O+SiO23. SiO2+C → CO2+Si¹ (SiO2)=100%-10%=90%m(SiO2)=0,9 ⋅ ]36 ]o]SiO 2 +C → CO2+ Si28 ]60 ]36 o ⋅ o]=60 28Hl\_lm(6L ]<ZjbZglDj_fgbc\ kh_^bg_gbyoijhy\ey_lkl_i_gvhdbke_gby±Si-4H + , Si+4O − , Si+2O-2[Z\2. Si → Mg2Si → SiH4 → SiO2+HCl→ Mg2SiZ6L0J [0J2Si+4HCl → 2MgCl2+SiH4\6L+4+2O2 → SiO2+2H2OtRt03.
2NaOH+SiO2 → Na2SiO3+H2O¹ (SiO2)=100%-10%=90%m(SiO2)=50 ⋅ ]45 ]tRo]2NaOH+ SiO 2 → Na 2SiO3 +H2O60 ]122 ]45o ⋅=o]60 122Hl\_lm(Na2SiO3 ]<ZjbZglHdkb^dj_fgby6L22 hlghkyld dbkehlgufhdkb^Zfih_]hk\hckl\ZfSiO2+2NaOH → Na2SiO3+H2OSiO2+CaO → CaSiO341Z[\2. Si → SiO2 → K2SiO3 → H2SiO3Z6L22 → SiO2[6L22+2KOH → K2SiO3+H2O\.2SiO3+2HCl → 2KCl+H2SiO3 ↓3. 2Mg+Si → Mg2Si¹ (Si)=100%-5%=95%m(Si)=0,95 ⋅ ]o]57 ]2Mg+ Si → Mg 2Si28 ]76 ]57 o ⋅ o]=28 76Hl\_lm(Mg26L ]JZ[hlZ<uqbke_gb_fZkkubebh[t_fZijh^mdlZj_Zdpbbihba\_klghcfZkk_bebihh[t_fmbkoh^gh]h\_s_kl\Zkh^_j`Zs_]hijbf_kb1.
CaCO3 → CaO + CO2ω(CaCO3) = 100% – 10% = 90%m(CaCO3) = 600 ⋅ 0,9 = 540 ]540 ]o]CaCO3 → CaO + CO2100 ]56 ]540x=100 56540 ⋅ 56x== 302,4 ]100Hl\_lm(CaO) = 302,4 ]2. CaCO3 → CaO + CO2ω(CaCO3) = 100% – 20% = 80%m(CaCO3) = 200 ⋅ 0,8 = 160 ]160 ]oe100 ]22,4 eCaCO3 → CaO + CO 2160x=100 22,4160 ⋅ 22,4x== 35,84 e100Hl\_l9&22) = 35,84 e3. N2 + 3H2 → 2NH342ω(N2) = 100% – 5% = 95%m(N2) = 50 ⋅ 0,95 = 47,5 ]47,5 ]o]28 ]2⋅17 ]N 2 + 3H2 → 2NH 347,5x=282 ⋅ 1747,5 ⋅ 2 ⋅ 17o]28η(NH3) =m ijm l_hj=8= 0,139 beb57, 7Hl\_lη(NH3) = 13,9%.t04. CaCO3 → CaO + CO2ω(CaCO3) = 100% – 6% = 94%m(CaCO3) = 0,94 ⋅ 400 = 376 d]376 d]o d]t0CaCO3 → CaO + CO2100 d]56 d]376x=100 5656 ⋅ 376od]100Hl\_lm(CaO) = 210,56 d]5.
CaCO3 → CaO + CO2ω(CaCO3) = 100% – 8% = 92%m(CaCO3) = 0,92 ⋅ 500 = 460 d]460 d]o d]of 3CaCO3 → CaO + CO 2100 d]56 d]22,4 f3460x=100 56460 ⋅ 56od]100460x=100 22,4460 ⋅ 22,4of3100Hl\_lm(CaO) = 257,6 d]V(CO2) = 103 f3.6.43o]2,24 e106 ]22,4 eNa 2 CO3 + 2HNO3 → 2NaNO3 + H2O + CO 2x2,24=106 22,4106 ⋅ 2,24o]22,410,6ω(Na2CO3) == 0,9815 beb10,8ωijbf = 100% – 98,15% = 1,85%Hl\_lωijbf = 1,85%.7.
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