ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿Ð¡ËÕ´òµÄÖ÷Òª³É·ÖÊÇ̼ËáÇâÄÆ£¬ÆäÖÐÍùÍùº¬ÓÐÉÙÁ¿µÄÂÈ»¯ÄÆ£¬»¯Ñ§ÐËȤС×éµÄͬѧҪͨ¹ýʵÑéÀ´²â¶¨Ä³Æ·ÅÆСËÕ´òÑùÆ·ÖÐ̼ËáÇâÄƵÄÖÊÁ¿·ÖÊý¡£
(1)׼ȷ³ÆÈ¡9.0gСËÕ´òÑùÆ··ÅÈëÉÕ±ÖУ¬ÖðµÎ¼ÓÈëÖÊÁ¿·ÖÊýΪ5%µÄÏ¡ÑÎËáÖÁÇ¡ºÃ²»ÔÙ²úÉúÆøÅÝΪֹ£¬¹²ÏûºÄÏ¡ÑÎËá73.0g£¬ÉÕ±ÖÐûÓвÐÁô²»ÈÜÎï¡£¼ÆËãÑùÆ·ÖÐ̼ËáÇâÄƵÄÖÊÁ¿·ÖÊý¡£(д³ö¼ÆËã¹ý³Ì)____
(2)ÇëÄãÉè¼ÆÒ»¸öÓëÉÏÊöʵÑéÔÀíºÍ²Ù×÷·½·¨¾ù²»ÏàͬµÄʵÑ飬²â¶¨Ð¡ËÕ´òÑùÆ·ÖÐ̼ËáÇâÄƵÄÖÊÁ¿·ÖÊý_______¡£
¡¾´ð°¸¡¿93.3%׼ȷ³ÆÈ¡ngСËÕ´òÑùÆ··ÅÈëÊÔ¹ÜÖУ¬¼ÓÈÈÖÁ²»ÔÙ²úÉúÆøÌåΪֹ£¬×¼È·³ÆÁ¿Ê£Óà¹ÌÌ壨Na2CO3£©ÖÊÁ¿Îªmg£¬ÀûÓ÷½³Ìʽ¿É¼ÆËãÑùÆ·ÖÐ̼ËáÇâÄƵÄÖÊÁ¿¼°ÖÊÁ¿·ÖÊý¡££¨·½°¸ºÏÀí¼´¿É£©
¡¾½âÎö¡¿
Éè̼ËáÇâÄƵÄÖÊÁ¿Îªx
NaHCO3 £« HCl=== NaCl£«H2O£«CO2¡ü
84¡¡¡¡ 36.5
x¡¡¡¡¡¡ 73.0g¡Á5%
=
x£½8.4 g
ÑùÆ·ÖÐ̼ËáÄƵÄÖÊÁ¿·ÖÊýΪ ¡Á100%£½93.3%
´ð£ºÑùÆ·ÖÐ̼ËáÇâÄƵÄÖÊÁ¿·ÖÊýΪ93.3%¡£
£¨2£©×¼È·³ÆÈ¡ngСËÕ´òÑùÆ··ÅÈëÊÔ¹ÜÖУ¬¼ÓÈÈÖÁ²»ÔÙ²úÉúÆøÌåΪֹ£¬×¼È·³ÆÁ¿Ê£Óà¹ÌÌ壨Na2CO3£©ÖÊÁ¿Îªmg£¬ÀûÓ÷½³Ìʽ¿É¼ÆËãÑùÆ·ÖÐ̼ËáÇâÄƵÄÖÊÁ¿¼°ÖÊÁ¿·ÖÊý¡££¨·½°¸ºÏÀí¼´¿É£©
ÉèÑùÆ·ÖÐ̼ËáÇâÄƵÄÖÊÁ¿Îªx
2NaHCO3Na2CO3+H2O+CO2¡ü
168 18+44
x n-m
=
x=¡Á£¨n-m£©
СËÕ´òµÄÖÊÁ¿·ÖÊýΪ¡Á£¨n-m£©¡Ân¡Á100%
¡¾ÌâÄ¿¡¿Ä³²Ýľ»Ò(Ö÷Òª³É·ÖΪK2CO3£¬ÔÓÖʲ»ÈÜÓÚË®)ÑùÆ·£¬¿ÉÄÜ»¹Ìí¼ÓÁËK2SO4¡¢KCl¡£Ä³»¯Ñ§ÐËȤС×é½øÐÐÁËÈçÏÂʵÑ飺
(1)ÑùÆ·µÄÔ¤´¦Àí£º
¢Ù²Ù×÷aµÄÃû³ÆΪ__________________£¬ÆäÄ¿µÄÊÇ____________________
¢ÚÈÜÒºAÖÐÖ÷ÒªµÄÒõÀë×Ó£¬³ýCO32-Í⣬¿ÉÄÜ»¹º¬ÓÐ_______________(ÌîÀë×Ó·ûºÅ)
(2)ÑùÆ·³É·ÖµÄÈ·¶¨£ºÉè¼ÆʵÑé·½°¸£¬Íê³ÉÏÂÊö±í¸ñ£»ÏÞÑ¡ÊÔ¼Á£ºÏ¡ÑÎËᡢϡÏõËᡢϡÁòËá¡¢ÂÈ»¯±µÈÜÒº¡¢ÏõËá±µÈÜÒº¡¢ÏõËáÒøÈÜÒº
ʵÑé²Ù×÷ | Ô¤ÆÚÏÖÏóºÍ±ØÒª½áÂÛ |
²½Öè1£ºÈ¡ÈÜÒºAÉÙÁ¿ÓÚÊÔ¹ÜÖУ¬¼ÓÈë×ãÁ¿ _______________£¬³ä·ÖÕñµ´£¬¾²Ö㬹ýÂË¡£ | |
²½Öè2£ºÈ¡ÉÙÁ¿²½Öè1ËùµÃµÄÂËÒºÓÚÊÔ¹ÜÖУ¬____________ | ______£¬ËµÃ÷ÑùÆ·ÖÐδÌí¼ÓKCl |
²½Öè3£ºÈ¡ÉÙÁ¿²½Öè1ËùµÃµÄÂËÔüÓÚÊÔ¹ÜÖУ¬__________________ | ______£¬ËµÃ÷ÑùÆ·ÖÐÌí¼ÓÁËK2SO4 |