ÌâÄ¿ÄÚÈÝ

£¨È¡»Æ¶¹Á£´óСµÄÒ»¿é½ðÊôÄÆ£¬ÓÃÂËÖ½²Á¸ÉÆä±íÃæµÄúÓÍ£¬È»ºó¼ÓÈëµ½ÁòËáÍ­ÈÜÒºÖУ¬¹Û²ìʵÑéÏÖÏó²¢Ð´³ö»¯Ñ§·½³Ìʽ£º¢ÙÄÆÔÚÈÜÒºÖз´Ó¦µÄʵÑéÏÖÏó
¸¡ÔÚÒºÃæÉÏ£¬ÈÛ³ÉÉÁÁÁµÄСÇò£¬·¢³öÏìÉù£¬ÓÐÆøÅݲúÉú
¸¡ÔÚÒºÃæÉÏ£¬ÈÛ³ÉÉÁÁÁµÄСÇò£¬·¢³öÏìÉù£¬ÓÐÆøÅݲúÉú
£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
2Na+2H2O=2NaOH+H2¡ü
2Na+2H2O=2NaOH+H2¡ü
£®¢ÚÈÜÒºÖеÄʵÑéÏÖÏó
ÓÐÀ¶É«³Áµí²úÉú
ÓÐÀ¶É«³Áµí²úÉú
£®·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
2NaOH+CuSO4=Cu£¨OH£©2¡ý+Na2SO4
2NaOH+CuSO4=Cu£¨OH£©2¡ý+Na2SO4
£®
·ÖÎö£º½ðÊôÄÆºÍÑÎÈÜÒºµÄ·´Ó¦ÏÈÊǺÍÀïÃæµÄË®µÄ·´Ó¦Éú³ÉÇâÑõ»¯ÄƺÍÇâÆø£¬È»ºó¿´ÇâÑõ»¯ÄƺÍÑÎÖ®¼äÊÇ·ñ»á·¢Éú¸´·Ö½â·´Ó¦£®
½â´ð£º½â£º¢ÙÏòÁòËáÍ­ÈÜÒºÖÐͶÈëһСÁ£ÄÆ£¬ÏÈÊǽðÊôÄÆºÍË®µÄ·´Ó¦£¬
ÏÖÏóÊÇ£º¸¡ÔÚÒºÃæÉÏ£¬ÈÛ³ÉÉÁÁÁµÄСÇò£¬·¢³öÏìÉù£¬ÓÐÆøÅݲúÉú£»
¢ÚÆä´ÎÊÇÉú³ÉµÄÇâÑõ»¯ÄƺÍÁòËáÍ­Ö®¼äµÄ·´Ó¦£¬Éú³ÉÇâÑõ»¯Í­À¶É«³ÁµíµÄ¹ý³Ì£¬
·´Ó¦·Ö±ðΪ£º2Na+2H2O=2NaOH+H2¡ü£¬2NaOH+CuSO4=Cu£¨OH£©2¡ý+Na2SO4£¬
¹Ê´ð°¸Îª£º¢Ù¸¡ÔÚÒºÃæÉÏ£¬ÈÛ³ÉÉÁÁÁµÄСÇò£¬£¬·¢³öÏìÉù£¬ÓÐÆøÅݲúÉú£»2Na+2H2O=2NaOH+H2¡ü£»
¢ÚÓÐÀ¶É«³Áµí²úÉú£»2NaOH+CuSO4=Cu£¨OH£©2¡ý+Na2SO4£®
µãÆÀ£º±¾Ì⿼²éѧÉú½ðÊôÄÆµÄ»¯Ñ§ÐÔÖÊ£ººÍÑÎÈÜÒºµÄ·´Ó¦£¬×¢Òâ¹æÂÉ£º½ðÊôÄÆºÍÑÎÈÜÒºµÄ·´Ó¦ÏÈÊǺÍÀïÃæµÄË®µÄ·´Ó¦Éú³ÉÇâÑõ»¯ÄƺÍÇâÆø£¬È»ºó¿´ÇâÑõ»¯ÄƺÍÑÎÖ®¼äÊÇ·ñ»á·¢Éú¸´·Ö½â·´Ó¦£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

£¨18·Ö£©£¨1£©±ê×¼×´¿öÏ£¬44.8LµÄNH3ÆøÌåÖÊÁ¿Îª       g£¬ÆäÖк¬        ¡¡¡¡  ¸öµªÔ­×Ó£¬ÆäÖк¬          mol ÇâÔ­×Ó¡£
(2)±ê×¼×´¿öÏ£¬11.2LµÄH2RÆøÌåÖÊÁ¿Îª17g£¬ÔòH2RµÄĦ¶ûÖÊÁ¿ÊÇ        £¬µÈÖÊÁ¿µÄNH3ÓëH2RµÄÎïÖʵÄÁ¿±ÈΪ          £¬1.7g°±ÆøÓë        mol H2Oº¬Óеĵç×ÓÊýÏàµÈ¡£
(3)¹ýÑõ»¯ÄƼ¸ºõ¿ÉÓëËùÓеij£¼ûÆøÌ¬·Ç½ðÊôÑõ»¯Îï·´Ó¦¡£È磺
2Na2O2 + 2CO2 = 2Na2CO3 + O2  £»  Na2O2 + CO = Na2CO3
ÊÔ·Ö±ðд³öNa2O2ÓëSO2¡¢SO3·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡
¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡
£¨4£©È¡»Æ¶¹Á£´óСµÄÒ»¿é½ðÊôÄÆ£¬ÓÃÂËÖ½²Á¸ÉÆä±íÃæµÄúÓÍ£¬È»ºó¼ÓÈëµ½ÁòËáÍ­ÈÜÒºÖУ¬¹Û²ìʵÑéÏÖÏó²¢Ð´³öÏà¹ØµÄ»¯Ñ§·½³Ìʽ£º
¢ÙʵÑéÏÖÏó                                                           
                                                                   
¢ÚÓйصĻ¯Ñ§·½³ÌʽÊÇ                                                ¡£
 
                                                                    ¡£

£¨18·Ö£©£¨1£©±ê×¼×´¿öÏ£¬44.8LµÄNH3ÆøÌåÖÊÁ¿Îª        g£¬ÆäÖк¬         ¡¡¡¡   ¸öµªÔ­×Ó£¬ÆäÖк¬          mol ÇâÔ­×Ó¡£

(2)±ê×¼×´¿öÏ£¬11.2LµÄH2RÆøÌåÖÊÁ¿Îª17g£¬ÔòH2RµÄĦ¶ûÖÊÁ¿ÊÇ         £¬µÈÖÊÁ¿µÄNH3ÓëH2RµÄÎïÖʵÄÁ¿±ÈΪ           £¬1.7g°±ÆøÓë         mol H2Oº¬Óеĵç×ÓÊýÏàµÈ¡£

 (3)¹ýÑõ»¯ÄƼ¸ºõ¿ÉÓëËùÓеij£¼ûÆøÌ¬·Ç½ðÊôÑõ»¯Îï·´Ó¦¡£È磺

2Na2O2 + 2CO2 = 2Na2CO3 + O2   £»  Na2O2 + CO = Na2CO3

ÊÔ·Ö±ðд³öNa2O2ÓëSO2¡¢SO3·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º

   ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡

   ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡

£¨4£©È¡»Æ¶¹Á£´óСµÄÒ»¿é½ðÊôÄÆ£¬ÓÃÂËÖ½²Á¸ÉÆä±íÃæµÄúÓÍ£¬È»ºó¼ÓÈëµ½ÁòËáÍ­ÈÜÒºÖУ¬¹Û²ìʵÑéÏÖÏó²¢Ð´³öÏà¹ØµÄ»¯Ñ§·½³Ìʽ£º

¢ÙʵÑéÏÖÏó                                                           

                                                                    

¢ÚÓйصĻ¯Ñ§·½³ÌʽÊÇ                                                 ¡£

 

                                                                     ¡£

 

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø