ÌâÄ¿ÄÚÈÝ
£¨18·Ö£©Ä³»î¶¯¿Î³ÌС×éÄâÓÃ50mLNaOHÈÜÒºÎüÊÕCO2ÆøÌ壬ÖƱ¸Na2CO3ÈÜÒº¡£ÎªÁË·ÀֹͨÈëµÄCO2ÆøÌå¹ýÁ¿Éú³ÉNaHCO3£¬Éè¼ÆÁËÈçÏÂʵÑé²½Ö裺
a.È¡25 mL NaOHÈÜÒºÎüÊÕ¹ýÁ¿µÄCO2ÆøÌ壬ÖÁCO2ÆøÌå²»ÔÙÈܽ⣻
b.С»ðÖó·ÐÈÜÒº1¡«2 min£»
c.Ôڵõ½µÄÈÜÒºÖмÓÈëÁíÒ»°ë(25mL)NaOHÈÜÒº£¬Ê¹Æä³ä·Ö»ìºÏ·´Ó¦¡£
(1)´Ë·½°¸ÄÜÖƵýϴ¿¾»µÄNa2CO3£¬Ð´³öa¡¢cÁ½²½µÄ»¯Ñ§·´Ó¦·½³Ì____________________¡¢____________________________¡£
´Ë·½°¸µÚÒ»²½µÄʵÑé×°ÖÃÈçÏÂͼËùʾ£º
(2)¼ÓÈë·´Ó¦ÎïÇ°£¬ÈçºÎ¼ì²éÕû¸ö×°ÖõÄÆøÃÜÐÔ______________________________________¡£
(3)×°ÖÃBÖÐÊ¢·ÅµÄÊÔ¼ÁÊÇ______________£¬
×÷ÓÃÊÇ___________________ ____________¡£
(4)ÔÚʵÑéÊÒͨ³£ÖÆ·¨ÖУ¬×°ÖÃA»¹¿É×÷ΪÏÂÁÐ_____________ ÆøÌåµÄ·¢Éú×°ÖÃ(ÌîÐòºÅ)¡£
¢ÙCH2==CH2 ¢ÚH2S ¢ÛCH4 ¢ÜCH¡ÔCH ¢ÝH2
(5)ʵÑéÊÒÖÆÈ¡ÏÂÁÐÆøÌ壺¢ÙNH3£¬¢ÚCl2£¬¢ÛHCl£¬¢ÜH2S£¬¢ÝCH4£¬¢ÞCO£¬¢ßCO2£¬¢àO2ʱ£¬ÊôÓÚ±ØÐë½øÐÐβÆø´¦Àí£¬²¢ÄÜÓÃÏÂͼËùʾװÖýøÐд¦ÀíµÄ£¬½«ÆøÌåµÄÐòºÅÌîÈë×°ÖÃͼµÄÏ·½¿Õ¸ñÄÚ¡£
(6)ÒÑÖªËùÓÃNaOHÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪ40%£¬ÊÒÎÂϸÃÈÜÒºÃܶÈΪ1.44 g / mL£¬¼ÙÉ跴ӦǰºóÈÜÒºµÄÌå»ý²»±ä£¬²»¿¼ÂÇʵÑéÎó²î£¬¼ÆËãÓôËÖÖ·½·¨ÖƱ¸ËùµÃNa2CO3ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ____________________¡£
a.È¡25 mL NaOHÈÜÒºÎüÊÕ¹ýÁ¿µÄCO2ÆøÌ壬ÖÁCO2ÆøÌå²»ÔÙÈܽ⣻
b.С»ðÖó·ÐÈÜÒº1¡«2 min£»
c.Ôڵõ½µÄÈÜÒºÖмÓÈëÁíÒ»°ë(25mL)NaOHÈÜÒº£¬Ê¹Æä³ä·Ö»ìºÏ·´Ó¦¡£
(1)´Ë·½°¸ÄÜÖƵýϴ¿¾»µÄNa2CO3£¬Ð´³öa¡¢cÁ½²½µÄ»¯Ñ§·´Ó¦·½³Ì____________________¡¢____________________________¡£
´Ë·½°¸µÚÒ»²½µÄʵÑé×°ÖÃÈçÏÂͼËùʾ£º
(2)¼ÓÈë·´Ó¦ÎïÇ°£¬ÈçºÎ¼ì²éÕû¸ö×°ÖõÄÆøÃÜÐÔ______________________________________¡£
(3)×°ÖÃBÖÐÊ¢·ÅµÄÊÔ¼ÁÊÇ______________£¬
×÷ÓÃÊÇ___________________ ____________¡£
(4)ÔÚʵÑéÊÒͨ³£ÖÆ·¨ÖУ¬×°ÖÃA»¹¿É×÷ΪÏÂÁÐ_____________ ÆøÌåµÄ·¢Éú×°ÖÃ(ÌîÐòºÅ)¡£
¢ÙCH2==CH2 ¢ÚH2S ¢ÛCH4 ¢ÜCH¡ÔCH ¢ÝH2
(5)ʵÑéÊÒÖÆÈ¡ÏÂÁÐÆøÌ壺¢ÙNH3£¬¢ÚCl2£¬¢ÛHCl£¬¢ÜH2S£¬¢ÝCH4£¬¢ÞCO£¬¢ßCO2£¬¢àO2ʱ£¬ÊôÓÚ±ØÐë½øÐÐβÆø´¦Àí£¬²¢ÄÜÓÃÏÂͼËùʾװÖýøÐд¦ÀíµÄ£¬½«ÆøÌåµÄÐòºÅÌîÈë×°ÖÃͼµÄÏ·½¿Õ¸ñÄÚ¡£
(6)ÒÑÖªËùÓÃNaOHÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪ40%£¬ÊÒÎÂϸÃÈÜÒºÃܶÈΪ1.44 g / mL£¬¼ÙÉ跴ӦǰºóÈÜÒºµÄÌå»ý²»±ä£¬²»¿¼ÂÇʵÑéÎó²î£¬¼ÆËãÓôËÖÖ·½·¨ÖƱ¸ËùµÃNa2CO3ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ____________________¡£
(1) 2NaOH + CO2 ="=" NaHCO3 NaHCO3 + NaOH ="=" Na2CO3 + H2O
(2)·½·¨¢Ù£ºÓõ¯»É¼Ð¼ÐסA¡¢BÁ¬½Ó´¦£¬Ïȼì²éAµÄÆøÃÜÐÔ£ºÈû½ôÏðƤÈû£¬´Ó©¶·×¢ÈëÒ»¶¨Á¿µÄË®£¬Ê¹Â©¶·ÖеÄË®Ãæ¸ßÓÚ׶ÐÎÆ¿ÄÚµÄË®Ã棬ֹͣ¼ÓË®ºó£¬Â©¶·ÄÚÓë׶ÐÎÆ¿ÖеÄÒºÃæ²î±£³Ö²»±ä£¬ËµÃ÷×°Öò»Â©Æø¡£È»ºó¼ì²éBµÄÆøÃÜÐÔ£ºÏòÉÕ±ÖÐ×¢ÈëÉÙÁ¿Ë®£¬Ê¹µ¼¹Ü¿ÚÇÖÈëË®ÖУ¬Ë«ÊÖÎæס¹ã¿ÚƿƬ¿ÌÓÐÆøÅÝð³ö£¬ËÉ¿ªÊÖºó£¬ÓÐÉÙÁ¿Ë®½øÈëµ¼¹ÜÐγÉË®Öù£¬ËµÃ÷×°Öò»Â©Æø¡£
·½·¨¢Ú£ºÒ²¿ÉÒ»´Î¼ì²éA¡¢BµÄÆøÃÜÐÔ£ºÁ¬½ÓºÍÉÕ±¼äµÄÈ齺¹ÜÓÃֹˮ¼Ð¼Ðס¡£È»ºó´Ó©¶·×¢ÈëÒ»¶¨Á¿µÄË®£¬Ê¹Â©¶·ÖеÄË®Ãæ¸ßÓÚ׶ÐÎÆ¿ÄÚµÄË®Ã棬¹ýÒ»»á£¬¹Û²ì©¶·ÄÚÓë׶ÐÎÆ¿ÖеÄÒºÃæ²î£¬Èô±£³Ö²»±ä£¬ËµÃ÷×°Öò»Â©Æø¡£
(3) ±¥ºÍËáÇâÄÆÈÜÒº ÎüÊÕHClÆøÌå
(4) ¢Ú¢Ü¢Ý (5) ¢Ù¢Û ¢Ú¢Ü (6) 7.2 mol/L
(2)·½·¨¢Ù£ºÓõ¯»É¼Ð¼ÐסA¡¢BÁ¬½Ó´¦£¬Ïȼì²éAµÄÆøÃÜÐÔ£ºÈû½ôÏðƤÈû£¬´Ó©¶·×¢ÈëÒ»¶¨Á¿µÄË®£¬Ê¹Â©¶·ÖеÄË®Ãæ¸ßÓÚ׶ÐÎÆ¿ÄÚµÄË®Ã棬ֹͣ¼ÓË®ºó£¬Â©¶·ÄÚÓë׶ÐÎÆ¿ÖеÄÒºÃæ²î±£³Ö²»±ä£¬ËµÃ÷×°Öò»Â©Æø¡£È»ºó¼ì²éBµÄÆøÃÜÐÔ£ºÏòÉÕ±ÖÐ×¢ÈëÉÙÁ¿Ë®£¬Ê¹µ¼¹Ü¿ÚÇÖÈëË®ÖУ¬Ë«ÊÖÎæס¹ã¿ÚƿƬ¿ÌÓÐÆøÅÝð³ö£¬ËÉ¿ªÊÖºó£¬ÓÐÉÙÁ¿Ë®½øÈëµ¼¹ÜÐγÉË®Öù£¬ËµÃ÷×°Öò»Â©Æø¡£
·½·¨¢Ú£ºÒ²¿ÉÒ»´Î¼ì²éA¡¢BµÄÆøÃÜÐÔ£ºÁ¬½ÓºÍÉÕ±¼äµÄÈ齺¹ÜÓÃֹˮ¼Ð¼Ðס¡£È»ºó´Ó©¶·×¢ÈëÒ»¶¨Á¿µÄË®£¬Ê¹Â©¶·ÖеÄË®Ãæ¸ßÓÚ׶ÐÎÆ¿ÄÚµÄË®Ã棬¹ýÒ»»á£¬¹Û²ì©¶·ÄÚÓë׶ÐÎÆ¿ÖеÄÒºÃæ²î£¬Èô±£³Ö²»±ä£¬ËµÃ÷×°Öò»Â©Æø¡£
(3) ±¥ºÍËáÇâÄÆÈÜÒº ÎüÊÕHClÆøÌå
(4) ¢Ú¢Ü¢Ý (5) ¢Ù¢Û ¢Ú¢Ü (6) 7.2 mol/L
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿