Question

(1)¼×´¼¿É×÷ΪȼÁϵç³ØµÄÔ­ÁÏ¡£ÒÔCH4ºÍH2OΪԭÁÏ,ͨ¹ýÏÂÁз´Ó¦À´ÖƱ¸¼×´¼¡£

¢ñ:CH4 (g)+H2O(g)=CO(g)+3H2(g) ¦¤H=+206.0 kJ¡¤mol-1

¢ò:CO(g)+2H2(g)=CH3OH(g) ¦¤H=-129.0 kJ¡¤mol-1

CH4(g)ÓëH2O(g)·´Ó¦Éú³ÉCH3OH (g)ºÍH2(g)µÄÈÈ»¯Ñ§·½³ÌʽΪ ¡£

(2)¼×´¼¶ÔË®ÖÊ»áÔì³ÉÒ»¶¨µÄÎÛȾ,ÓÐÒ»Öֵ绯ѧ·¨¿ÉÏû³ýÕâÖÖÎÛȾ,ÆäÔ­ÀíÊÇ:ͨµçºó,½«Co2+Ñõ»¯³ÉCo3+,È»ºóÒÔCo3+×÷Ñõ»¯¼Á°ÑË®Öеļ״¼Ñõ»¯³ÉCO2¶ø¾»»¯¡£ÊµÑéÊÒÓÃÈçͼװÖÃʵÏÖÉÏÊö¹ý³Ì:

¢Ùд³öÑô¼«µç¼«·´Ó¦Ê½: ¡£

¢Úд³ö³ýÈ¥¼×´¼µÄÀë×Ó·½³Ìʽ:  ¡£

(3)д³öÒÔNaHCO3ÈÜҺΪ½éÖʵÄAl¡ª¿ÕÆøÔ­µç³ØµÄ¸º¼«·´Ó¦Ê½: ¡£

(4)È˹¤ÉöÔà¿É²ÉÓüä½Óµç»¯Ñ§·½·¨³ýÈ¥´úл²úÎïÖеÄÄòËØ,Ô­ÀíÈçͼ:

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Answer 1

(1)CH4(g)+H2O(g)=CH3OH (g)+H2(g) ¦¤H=+77.0 kJ¡¤mol-1

(2)¢ÙCo2+-e-=Co3+

¢Ú6Co3++CH3OH+H2O=CO2¡ü+6Co2++6H+

(3) Al-3e-+3HC=Al(OH)3¡ý+3CO2¡ü

(4)¢ÙB ¢Ú2Cl--2e-=Cl2¡ü¡¢CO(NH2)2+3Cl2+H2O=CO2+N2+6HCl

¢Û²»±ä 7.2

¡¾½âÎö¡¿(1)¢ñ:CH4(g)+H2O(g)=CO(g)+3H2(g) ¦¤H=+206.0 kJ¡¤mol-1

¢ò:CO(g)+2H2(g)=CH3OH(g) ¦¤H=-129.0 kJ¡¤mol-1

¢ñ+¢òµÃ:CH4(g)+H2O(g)=CH3OH(g)+H2(g) ¦¤H=+77.0 kJ¡¤mol-1

(2)¸ù¾Ý·´Ó¦Ô­Àí,ͨµçºó,½«Co2+Ñõ»¯³ÉCo3+,È»ºóÒÔCo3+×÷Ñõ»¯¼Á°ÑË®Öеļ״¼Ñõ»¯³ÉCO2¶ø¾»»¯,ÔòͨµçʱCo2+ʧµç×Ó,Ñô¼«µÄµç¼«·´Ó¦Ê½ÎªCo2+-e-=Co3+¡£³ýÈ¥¼×´¼µÄÀë×Ó·½³Ìʽ6Co3++CH3OH+H2O=CO2¡ü+6Co2++6H+¡£

(3)Al¡ª¿ÕÆøÔ­µç³ØÂÁ×÷¸º¼«,ʧµç×ÓÉú³ÉÂÁÀë×Ó,ºÍ̼ËáÇâÄÆ·´Ó¦Éú³ÉÇâÑõ»¯ÂÁºÍ¶þÑõ»¯Ì¼¡£¸º¼«:Al-3e-+3HC=Al(OH)3¡ý+3CO2¡ü¡£

(4)¢ÙÓÉͼ¿ÉÖª²úÉúCl2µÄµç¼«ÎªÑô¼«,¹ÊAΪÕý¼«,BΪ¸º¼«¡£¢ÚÑô¼«ÊÒ·¢ÉúµÄ·´Ó¦Ê×ÏÈΪ2Cl--2e-=Cl2¡ü¡£ÓɲúÎïCO2¡¢N2¿ÉÖªCO(NH2)2ÔÚ´ËÊÒ±»Ñõ»¯,Cl2±»»¹Ô­,¹Ê·´Ó¦ÎªCO (NH2)2+3Cl2+H2O=CO2+N2+6HCl¡£¢ÛÒõ¼«·¢Éú2H++2e-=H2¡ü,ͬʱÑô¼«ÊÒHClµçÀë³öµÄH+ͨ¹ýÖÊ×Ó½»»»Ä¤½øÈëÒõ¼«ÊÒ,´Ó¶øʹÒõ¼«ÊÒH+Ũ¶È±£³Ö²»±ä,¼´pHÓëµç½âÇ°Ïà±È²»±ä¡£Á½¼«¹²ÊÕ¼¯µ½µÄÆøÌån(ÆøÌå)==0.6 mol,ÓÉ:

CO(NH2)2¡«3Cl2¡«3H2¡«CO2¡«N2

60 g 5 mol

m[CO(NH2)2] 0.6 mol

µÃ:m[CO(NH2)2]==7.2 g¡£