ÌâÄ¿ÄÚÈÝ
Ïò100 mL NaOHÈÜÒºÖÐͨÈëCO2³ä·Ö·´Ó¦ºó£¬ÔÚ¼õѹºÍ½ÏµÍζÈÏ£¬Ð¡Ðĵؽ«ÈÜÒºÕô¸É£¬µÃµ½°×É«¹ÌÌåM¡£Í¨Èë¶þÑõ»¯Ì¼µÄÌå»ýV(±ê×¼×´¿öÏÂ)ÓëMµÄÖÊÁ¿WµÄ¹ØϵÈçͼ
Ëùʾ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)bµãʱMµÄ×é³É³É·ÖΪ______________________¡£
(2)ÈôҪʹbµãÉú³ÉµÄÑεÄÖÊÁ¿±äΪ8.4 g£¬ÔòÓ¦¼ÌÐøÏòÈÜÒºÖÐͨÈë¶þÑõ»¯Ì¼________L(±ê×¼×´¿öÏÂ)¡£
(3)ÈôÏòÉú³ÉµÄ7.16 gÑεÄÈÜÒºÖмÓÈëÒ»¶¨Á¿µÄijÎïÖÊ£¬³ä·Ö·´Ó¦ºó£¬¼õѹµÍÎÂÕô·¢µÃµ½´¿¾»µÄ̼ËáÄƹÌÌå(Î޽ᾧˮ)8.4 g¡£
¢ÙÈôÖ»¼ÓÈë0.03 molijÎïÖÊ£¬Ôò¼ÓÈëµÄÎïÖÊ¿ÉÒÔÊÇ________»ò________£»
¢ÚÈôÖ»¼ÓÈë0.06 molijÎïÖÊ£¬Ôò¼ÓÈëµÄÎïÖÊ¿ÉÒÔÊÇ________¡¢________»ò________¡£
(4)³£ÎÂÏ£¬Í¬Å¨¶ÈµÄ̼ËáÄÆÈÜÒººÍ̼ËáÇâÄÆÈÜÒºµÄpH
¶¼´óÓÚ7£¬¶þÕßÖÐ________µÄpH¸ü´ó£¬ÀíÓÉÊÇ________________________£»0.1 mol¡¤L£1̼ËáÄÆÈÜÒºÖÐÀë×ÓŨ¶ÈµÄ´óС¹ØϵÊÇ________£»Ïò̼ËáÇâÄÆÈÜÒºÖÐÖðµÎµÎÈëÇâÑõ»¯±µÈÜÒº£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ________¡£
Ëùʾ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)bµãʱMµÄ×é³É³É·ÖΪ______________________¡£
(2)ÈôҪʹbµãÉú³ÉµÄÑεÄÖÊÁ¿±äΪ8.4 g£¬ÔòÓ¦¼ÌÐøÏòÈÜÒºÖÐͨÈë¶þÑõ»¯Ì¼________L(±ê×¼×´¿öÏÂ)¡£
(3)ÈôÏòÉú³ÉµÄ7.16 gÑεÄÈÜÒºÖмÓÈëÒ»¶¨Á¿µÄijÎïÖÊ£¬³ä·Ö·´Ó¦ºó£¬¼õѹµÍÎÂÕô·¢µÃµ½´¿¾»µÄ̼ËáÄƹÌÌå(Î޽ᾧˮ)8.4 g¡£
¢ÙÈôÖ»¼ÓÈë0.03 molijÎïÖÊ£¬Ôò¼ÓÈëµÄÎïÖÊ¿ÉÒÔÊÇ________»ò________£»
¢ÚÈôÖ»¼ÓÈë0.06 molijÎïÖÊ£¬Ôò¼ÓÈëµÄÎïÖÊ¿ÉÒÔÊÇ________¡¢________»ò________¡£
(4)³£ÎÂÏ£¬Í¬Å¨¶ÈµÄ̼ËáÄÆÈÜÒººÍ̼ËáÇâÄÆÈÜÒºµÄpH
¶¼´óÓÚ7£¬¶þÕßÖÐ________µÄpH¸ü´ó£¬ÀíÓÉÊÇ________________________£»0.1 mol¡¤L£1̼ËáÄÆÈÜÒºÖÐÀë×ÓŨ¶ÈµÄ´óС¹ØϵÊÇ________£»Ïò̼ËáÇâÄÆÈÜÒºÖÐÖðµÎµÎÈëÇâÑõ»¯±µÈÜÒº£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ________¡£
(1)Na2CO3ºÍNaHCO3¡¡(2)0.448
(3)¢ÙNa2O¡¡Na2O2¡¡¢ÚNa¡¡NaOH¡¡NaH
(4)Na2CO3¡¡ÔÚÏàͬÌõ¼þÏ£¬CO32¡ªµÄË®½âÄÜÁ¦±ÈHCO3¡ªµÄË®½âÄÜÁ¦Ç¿¡¡c(Na£«)£¾c(CO32¡ª)£¾c(OH£)£¾c(HCO3¡ª)£¾c(H£«)¡¡Ba2£«£«2OH££«2HCO3¡ª=BaCO3¡ý£«CO32¡ª£«2H2O
(3)¢ÙNa2O¡¡Na2O2¡¡¢ÚNa¡¡NaOH¡¡NaH
(4)Na2CO3¡¡ÔÚÏàͬÌõ¼þÏ£¬CO32¡ªµÄË®½âÄÜÁ¦±ÈHCO3¡ªµÄË®½âÄÜÁ¦Ç¿¡¡c(Na£«)£¾c(CO32¡ª)£¾c(OH£)£¾c(HCO3¡ª)£¾c(H£«)¡¡Ba2£«£«2OH££«2HCO3¡ª=BaCO3¡ý£«CO32¡ª£«2H2O
(1)ÓÉͼÏóÖª£¬¿ªÊ¼Ê±ÇâÑõ»¯ÄÆΪ4.0 g£¬ÎïÖʵÄÁ¿Îª0.1 mol£¬ÈôÈ«²¿Éú³É̼ËáÄÆ£¬ÔòWΪ5.3 g£»ÈôÈ«²¿Éú³É̼ËáÇâÄÆ£¬ÔòWΪ8.4 g£¬bµã¶ÔÓ¦µÄ¹ÌÌåMΪ7.16 g£¬½éÓÚ5.3 gºÍ8.4 gÖ®¼ä¡£¹Êbµã¶ÔÓ¦µÄ¹ÌÌåΪ̼ËáÄƺÍ̼ËáÇâÄƵĻìºÏÎï¡£(2)Éèbµã¶ÔÓ¦µÄ¹ÌÌåÖÐ̼ËáÄÆ¡¢Ì¼ËáÇâÄƵÄÎïÖʵÄÁ¿·Ö±ðΪx¡¢y£¬ÔòÓÐ106 g¡¤mol£1¡Áx£«84 g¡¤mol£1¡Áy£½7.16 g,2x£«y£½0.1 mol£¬½âµÃ£ºx£½0.02 mol£¬y£½0.06 mol¡£ÓÉNa2CO3£«H2O£«CO2=2NaHCO3µÃ£¬n(CO2)£½0.02 mol¡£(3)8.4 g Na2CO3µÄÎïÖʵÄÁ¿Îª0.08 mol£¬ÒÀÌâÒ⣬0.06 mol NaHCO3¨D¡ú0.06 mol Na2CO3£¬Ôò·¢ÉúµÄ·´Ó¦ÎªNaHCO3£«NaOH=Na2CO3£«H2O£¬n(NaOH)£½0.06 mol£¬ÈÜÓÚË®²úÉúÇâÑõ»¯ÄƵÄÎïÖÊÖУ¬ÄÆ¡¢¹ýÑõ»¯ÄÆ¡¢Ñõ»¯ÄÆ¡¢Ç⻯ÄÆ(NaH£«H2O=NaOH£«H2¡ü)²»ÒýÈëÐÂÔÓÖÊ¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿