СËÕ´ò(NaHCO3)ÊÇʳƷ¹¤ÒµÖÐÒ»ÖÖÓ¦ÓÃ×î¹ã·ºµÄÌí¼Ó¼Á¡£×ÊÁÏÏÔʾ£¬NaHCO3ÔÚÉú²ú¹ý³ÌÖпÉÄܲ¿·Ö·Ö½â³ÉNa2CO3¡£Ä³»¯Ñ§ÐËȤС×éͬѧ³ÆÈ¡28£®95g¹¤ÒµNaHCO3ÑùÆ·½øÐзÖÎö£¬ÏÈÓþƾ«µÆ¶ÔÆä³ä·Ö¼ÓÈÈ£¬ÀäÈ´ºó³ÆµÃÆäÖÊÁ¿Îª21£®2g£»Áí³ÆÈ¡Ò»·ÝÏàͬÖÊÁ¿µÄÑùÆ·£¬ÖðµÎ¼ÓÈëÏ¡ÁòËáÖÁûÓÐÆøÅݲúÉúΪֹ£¬¹²ÊÕ¼¯µ½ÆøÌåÕÛËã³É±ê×¼×´¿öϵÄÌå»ýΪ7£®28L¡££¨¼ÙÉèÑùÆ·ÖгýÁËNa2CO3ºÍNaHCO3Í⣬ÈçÓÐÆäËüÔÓÖÊ£¬ÔÓÖÊÊÜÈÈÎȶ¨ÇÒ²»ÓëÁòËá·´Ó¦£©Í¨¹ý¼ÆËã»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öNaHCO3ÊÜÈÈ·Ö½âµÄ»¯Ñ§·´Ó¦·½³Ìʽ_________________________£»
£¨2£©¼ÓÈȺóµÃµ½µÄ21£®2g¹ÌÌåÊÇ____________£¨Ìѧʽ£©£»
£¨3£©ÑùÆ·ÖÐNa2CO3µÄÖÊÁ¿·ÖÊýÊǶàÉÙ£¨Ð´³ö¼ÆËã¹ý³Ì£©¡£
£¨1£©2NaHCO3Na2CO3+CO2¡ü+H2
£¨2£©Na2CO3
£¨3£©ÉèÑùÆ·ÖÐNa2CO3¡¢NaHCO3µÄÎïÖʵÄÁ¿·Ö±ðΪx¡¢y
Na2CO3 +H2SO4 £½ Na2SO4 +CO2¡ü+H2O
2NaHCO3 + H2SO4 =Na2SO4+2CO2¡ü+2H2O

x=0.075mol
ÑùÆ·ÖÐNa2CO3µÄÖÊÁ¿·ÖÊýΪ£º0.075mol¡Á106g/mol¡Â28.95g¡Á100%=27.46©‡
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁи÷À¸¶ÔÓ¦¹Øϵȫ²¿ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
Ñ¡Ïî Ë×³Æ Ö÷Òª³É·Ö»¯Ñ§Ê½ Àà±ð
A ´¿¼î NaOH ¼î
B СËÕ´ò NaHCO3 Ëá
C ´ÅÌú¿ó Fe2O3 ½ðÊôÑõ»¯Îï
D ÂÁÍÁ¿ó Al2O3 Á½ÐÔÑõ»¯Îï

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÇé¿ö»á¶ÔÈËÌ彡¿µÔì³É½Ï´óΣº¦µÄÊÇ(    )

A.×ÔÀ´Ë®ÖÐͨÈëÉÙÁ¿Cl2½øÐÐÏû¶¾É±¾ú

B.ÓÃÃ×´×ÇåÏ´ÈÈˮƿµ¨ÄÚ±Ú¸½×ŵÄË®¹¸(º¬CaCO3)

C.ÓÃSO2Ư°×ʳƷ

D.ÓÃСËÕ´ò(NaHCO3)·¢½ÍÃæÍÅ£¬ÖÆ×÷ÂøÍ·

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Ò½Ò©°¢Ë¹Æ¥ÁֵĽṹ¼òʽÈçͼ¡£ÊԻشð£º

£¨1£©°¢Ë¹Æ¥ÁÖ¿É¿´³Éõ¥ÀàÎïÖÊ£¬¿Ú·þºóÔÚ賦×÷ÓÃÏ£¬°¢Ë¹Æ¥ÁÖ·¢ÉúË®½â·´Ó¦£¬Éú³ÉAºÍBÁ½ÖÖ²úÎï¡£ÆäÖÐAµÄ½á¹¹¼òʽÈçͼ£¬ÔòBµÄ½á¹¹¼òʽΪ£º              £¬AÖеĺ¬Ñõ¹ÙÄÜÍÅÃû³ÆÊÇ£º         ¡¢        ¡£

£¨2£©°¢Ë¹Æ¥ÁÖ¸úСËÕ´ò(NaHCO3)ͬʱ·þÓ㬿ÉʹÉÏÊöË®½â²úÎïAÓëСËÕ´ò·´Ó¦£¬Éú³É¿ÉÈÜÐÔÑÎËæÄòÒºÅųö£¬´ËÑεĽṹ¼òʽΪ£º                     ¡£

£¨3£©ÉÏÊöË®½â²úÎïAÓëNaOHÈÜÒº·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º

                                                              ¡£

£¨4£©ÉÏÊöË®½â²úÎïA ÓëŨäåË®·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º

                                                              ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Ò½Ò©°¢Ë¹Æ¥ÁֵĽṹ¼òʽÈçͼ¡£ÊԻشð£º

£¨1£©°¢Ë¹Æ¥ÁÖ¿É¿´³Éõ¥ÀàÎïÖÊ£¬¿Ú·þºóÔÚ賦×÷ÓÃÏ£¬°¢Ë¹Æ¥ÁÖ·¢ÉúË®½â·´Ó¦£¬Éú³ÉAºÍBÁ½ÖÖ²úÎï¡£ÆäÖÐAµÄ½á¹¹¼òʽÈçͼ£¬ÔòBµÄ½á¹¹¼òʽΪ£º              £¬AÖеĺ¬Ñõ¹ÙÄÜÍÅÃû³ÆÊÇ£º         ¡¢        ¡£

£¨2£©°¢Ë¹Æ¥ÁÖ¸úСËÕ´ò(NaHCO3)ͬʱ·þÓ㬿ÉʹÉÏÊöË®½â²úÎïAÓëСËÕ´ò·´Ó¦£¬Éú³É¿ÉÈÜÐÔÑÎËæÄòÒºÅųö£¬´ËÑεĽṹ¼òʽΪ£º                     ¡£

£¨3£©ÉÏÊöË®½â²úÎïAÓëNaOHÈÜÒº·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º

                                                              ¡£

£¨4£©ÉÏÊöË®½â²úÎïA ÓëŨäåË®·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º

                                                              ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ì½­ËÕÊ¡¸ß¶þÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í£¨±ØÐÞ£© ÌâÐÍ£ºÌî¿ÕÌâ

±£³ÖÓªÑøƽºâ£¬ºÏÀíʹÓÃÒ©ÎïÊDZ£Ö¤ÉíÐĽ¡¿µ¡¢Ìá¸ßÉú»îÖÊÁ¿µÄÓÐЧÊֶΡ£

£¨1£©£¨2·Ö£© ¢ÙÈËÀàµÄÉúÃü»î¶¯ÐèÒªÌÇÀà¡¢             ¡¢             Î¬ÉúËØ¡¢Ë®¡¢ºÍÎÞ»úÑÎ(»ò¿óÎïÖÊ)µÈÁù´óËØÓªÑøÎïÖÊ¡£

 ¢Ú£¨3·Ö£©°¢Ë¾Æ¥ÁÖ¾ßÓР          ×÷Óᣳ¤ÆÚ´óÁ¿·þÓð¢Ë¾Æ¥ÁÖ£¬ÆäË®½â²úÎïË®ÑîËá¿Éµ¼Ö»¼Õß³öÏÖÍ·Í´¡¢¶ñÐĵÈÖ¢×´£¬Ðè¾²Âö×¢ÉäСËÕ´ò(NaHCO3)ÈÜÒº£¬ÀûÓÃСËÕ´òÓëË®ÑîËá·Ö×ÓÖеÄôÈ»ù·´Ó¦Éú³ÉË®ÑîËáÄÆ£¬Ê¹Ö¢×´»º½â¡£Ð´³öË®ÑîËáÓëСËÕ´ò·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º                         ¡£

£¨2£©£¨2·Ö£©µ°°×ÖÊÔÚÈËÌåÄÚË®½âµÄ×îÖÕ²úÎïÊÇ°±»ùËá¡£ÇëÔÚÏÂͼÐéÏß·½¿òÄÚдÉÏÊʵ±µÄ¹ÙÄÜÍÅ·ûºÅ£¬½«°±»ùËáµÄͨʽ²¹³äÍêÕû£º

                                                                                             

£¨3£©µí·ÛË®½âµÄ×îÖÕ²úÎïÊÇ          ÈôÒª¼ìÑéµí·ÛµÄµí·Ûø×÷ÓÃÏÂÒѾ­·¢ÉúÁËË®½â£¬¿ÉÈ¡ÉÙÁ¿ÉÏÊöÈÜÒº¼ÓÈë¼îÒºµ÷µ½¼îÐÔ£¬ÔÙ¼ÓÈë            £¨ÌîÊÔ¼ÁµÄÃû³Æ£©£¬¼ÓÈȺóÔÙ¸ù¾ÝʵÑéÏÖÏóÅжϣ»ÈôÒª¼ìÑéµí·ÛûÓÐÍêÈ«Ë®½âµÄ£¬¿ÉÈ¡ÉÙÁ¿ÉÏÊöÈÜÒº¼ÓÈ뼸µÎ         ÈÜÒº£¬Ó¦¹Û²ìµ½³öÏÖÀ¶É«¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸