ΪÁËÔ¤·À»·¾³ÎÛȾ²¢¶ÔβÆø½øÐÐ×ÛºÏÀûÓã¬Ä³ÁòË᳧Óð±Ë®ÎüÊÕβÆøÖеÄSO2¡£ÔÙÏòÎüÊÕÒºÖмÓÈëŨÁòËᣬÒÔÖÆÈ¡¸ßŨ¶ÈµÄSO2¼°£¨NH4£©2SO4ºÍNH4HSO4¹ÌÌå¡£

Ϊ²â¶¨ÉÏÊö£¨NH4£©2SO4ºÍNH4HSO4¹ÌÌå»ìºÏÎïµÄ×é³É£¬ÏÖ³ÆÈ¡¸ÃÑùÆ·4·Ý£¬·Ö±ð¼ÓÈëÏàͬŨ¶ÈµÄNaOHÈÜÒº¸÷40.00 mL£¬¼ÓÈÈÖÁ120 ¡æ×óÓÒ£¬Ê¹°±ÆøÈ«²¿Òݳö¡²£¨NH4£©2SO4ºÍNH4HSO4¹ÌÌå·Ö½âµÄζȾù¸ßÓÚ200 ¡æ¡³£¬²âµÃÓйØʵÑéÊý¾ÝÈçÏ£º£¨±ê×¼×´¿öÏ£©

ʵÑéÐòºÅ

ÑùÆ·µÄÖÊÁ¿/g

NaOHÈÜÒºµÄÌå»ý/mL

°±ÆøµÄÌå»ý/L

¢ñ

7.4

40.00

1.68

¢ò

14.8

40.00

3.36

¢ó

22.2

40.00

1.12

¢ô

37.0

40.00

0.00

£¨1£©ÊµÑé¹ý³ÌÖÐÓйط´Ó¦µÄÀë×Ó·½³ÌʽΪ_________________________________________¡£

£¨2£©ÓÉ¢ñ×éÊý¾ÝÖ±½ÓÍƲ⣺±ê×¼×´¿öÏÂ3.7 g ÑùÆ·½øÐÐͬÑùʵÑéʱ£¬Éú³É°±ÆøµÄÌå»ýΪ__________L¡£

£¨3£©ÊÔ¼ÆËã¸Ã»ìºÏÎïÖУ¨NH4£©2SO4ºÍNH4HSO4µÄÎïÖʵÄÁ¿Ö®±ÈΪ__________________¡£

£¨4£©Óû¼ÆËã¸ÃNaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÓ¦¸ÃÑ¡Ôñ__________×éÊý¾Ý£¬ÓÉ´ËÇóµÃNaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ____________________________________________________________¡£

£¨1£©H++OH-====H2O£¬+OH-NH3¡ü+H2O  £¨2£©0.84  (3)1¡Ã4  (4)¢ó  5 mol¡¤L-1


½âÎö:

£¨1£©£¨NH4£©2SO4ºÍNH4HSO4ÔÚÈÜÒºÖÐÊÇÍêÈ«µçÀëµÄ£¬ËùÒÔ£¨NH4£©2SO4ºÍNH4HSO4¹ÌÌå»ìºÏÎïÖмÓÈëNaOHÈÜҺʱ£¬½«ÓÐÒÔÏÂÁ½¸öÀë×Ó·´Ó¦·¢ÉúH++OH-====H2O£¬+OH-NH3¡ü+H2O¡£

£¨2£©ÓɱíÖÐÊý¾Ý¿É¿´³ö£¬µ±ÑùÆ·ÖÊÁ¿ÓÉ7.4 gÔö¼Óµ½14.8 gʱ£¬ÊͷųöµÄÆøÌåÌå»ýÓÉ1.68 LÔö¼Óµ½3.36 L£¬ÊdzɱÈÀýÔö¼ÓµÄ£¬ËµÃ÷NaOHÈÜÒº´ËʱÊÇ×ãÁ¿µÄ¡£ËùÒÔ¿ÉÍƶϳöµ±ÑùÆ·µÄÖÊÁ¿Îª3.7 gʱ£¬ËùÊͷųöµÄÆøÌåÌå»ýӦΪ1.68 L/2=0.84 L¡£

£¨3£©ÉèI×é»ìºÏÎïÖУ¨NH4£©2SO4µÄÎïÖʵÄÁ¿Îªx mol£¬NH4HSO4µÄÎïÖʵÄÁ¿Îªy mol£¬ÒÀ¾Ý±íÖеÄÊý¾ÝÓУºµ±»ìºÏÎïÖÊÁ¿Îª14.8 gʱ£¬²úÉúÆøÌåNH3µÄÎïÖʵÄÁ¿Îª3.36 L/22.4 L¡¤mol-1=0.15 mol¡£

£¨NH4£©2SO4¡ª2NH3      NH4HSO4¡ªNH3

  x mol       2x mol     y mol      y mol

  ½â´Ë·½³Ì×éµÃµ½£ºx=0.025,y=0.1¡£

ËùÒÔ£¬»ìºÏÎïÖУ¨NH4£©2SO4ºÍNH4HSO4µÄÎïÖʵÄÁ¿Ö®±ÈΪ0.025¡Ã0.1=1¡Ã4¡£

(4)·ÖÎö±íÖÐÊý¾Ý£¬µ±¼ÓÈëÑùÆ·ÖÊÁ¿Îª22.2 gʱ£¬·Å³öµÄÆøÌåÌå»ý²»ÔÙÓëÑùÆ·ÖÊÁ¿³É±ÈÀý£¬ËµÃ÷´ËʱNaOHµÄÁ¿²»×㣬µ«¿ÉÒÔÅжϳö£¬´ËʱµÄNaOHÏûºÄÍêÁËÈ«²¿µÄH+£¬²¢ÏûºÄÁ˲¿·ÖµÄ¡£µ±¼ÓÈëÑùƷΪ37.0 gʱ£¬ÎÞÆøÌåÊÍ·Å£¬Õâ±íÃ÷´ËʱNaOHÈ«²¿±»ÈÜÒºÖеÄH+ÏûºÄ¡£¼ÆËã¸ÃNaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÓ¦¸ÃÑ¡Ôñ¢ó×éÊý¾Ý½øÐмÆËã¡£

½áºÏ£¨3£©µÄ¼ÆËã½á¹û¿ÉÖª£¬ÔÚ22.2 gÑùÆ·Öк¬£¨NH4£©2SO4ӦΪ0.025 mol¡Á1.5=0.037 5 mol£¬º¬ÓÐNH4HSO4µÄÎïÖʵÄÁ¿Îª0.1 mol¡Á1.5=0.15 mol¡£

ÓÉÓÚH+¡ªNaOH£¬ÔòÈÜÒºÖеÄH+ÏûºÄµÄNaOHΪ0.15 mol¡£

ÓÉÓÚ¡ªNaOH¡ªNH3,ÈÜÒºÖеÄÏûºÄµÄNaOHΪ1.12 L/2.24 L¡¤mol-1=0.05 mol¡£

ËùÒÔ£¬NaOHµÄ×ÜÎïÖʵÄÁ¿Îª0.15 mol+0.05 mol=0.2 mol¡£

NaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ0.2 mol/0.04 L=5 mol¡¤L-1¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ºÖØÄѵãÊֲᡡ¸ßÖл¯Ñ§¡¤±ØÐÞ1¡¡ÅäÈ˽̰æпαê È˽̰æпαê ÌâÐÍ£º022

ΪÁËÔ¤·À»·¾³ÎÛȾ²¢¶ÔβÆø½øÐÐ×ÛºÏÀûÓã¬Ä³ÁòË᳧Óð±Ë®ÎüÊÕβÆøÖеÄSO2£®ÔÙÏòÎüÊÕÒºÖмÓÈëŨÁòËᣬÒÔÖÆÈ¡¸ßŨ¶ÈµÄSO2¼°(NH4)2SO4ºÍNH4HSO4¹ÌÌ壮

Ϊ²â¶¨ÉÏÊö(NH4)2SO4ºÍNH4HSO4¹ÌÌå»ìºÏÎïµÄ×é³É£¬ÏÖ³ÆÈ¡¸ÃÑùÆ·4·Ý£¬·Ö±ð¼ÓÈëÏàͬŨ¶ÈµÄNaOHÈÜÒº¸÷40.00 mL£¬¼ÓÈÈÖÁ120¡æ×óÓÒ£¬Ê¹°±È«²¿Òݳö[(NH4)2SO4ºÍNH4HSO4¹ÌÌå·Ö½âµÄζȾù¸ßÓÚ200¡æ]£¬²âµÃÓйØʵÑéÊý¾ÝÈçÏÂ(±ê×¼×´¿öÏÂ)£º

(1)ʵÑé¹ý³ÌÖÐÓйط´Ó¦µÄÀë×Ó·½³ÌʽΪ________£®

(2)ÓÉ¢ñ×éÊý¾ÝÖ±½ÓÔ¤²â£º±ê×¼×´¿öÏÂ3.7 gÑùÆ·½øÐÐͬÑùʵÑéʱ£¬Éú³É°±µÄÌå»ýΪ________L£®

(3)¸Ã»ìºÏÎïÖÐ(NH4)2SO4ºÍNH4HSO4µÄÎïÖʵÄÁ¿Ö®±ÈΪ________£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ΪÁËÔ¤·À»·¾³ÎÛȾ²¢¶ÔβÆø½øÐÐ×ÛºÏÀûÓã¬Ä³ÁòË᳧Óð±Ë®ÎüÊÕβÆøÖеÄSO2¡£ÔÙÏòÎüÊÕÒºÖмÓÈëŨÁòËᣬÒÔÖÆÈ¡¸ßŨ¶ÈµÄSO2¼°£¨NH4£©2SO4ºÍNH4HSO4¹ÌÌå¡£

Ϊ²â¶¨ÉÏÊö£¨NH4£©2SO4ºÍNH4HSO4¹ÌÌå»ìºÏÎïµÄ×é³É£¬ÏÖ³ÆÈ¡¸ÃÑùÆ·4·Ý£¬·Ö±ð¼ÓÈëÏàͬŨ¶ÈµÄNaOHÈÜÒº¸÷40.00 mL£¬¼ÓÈÈÖÁ120 ¡æ×óÓÒ£¬Ê¹°±ÆøÈ«²¿Òݳö¡²£¨NH4£©2SO4ºÍNH4HSO4¹ÌÌå·Ö½âµÄζȾù¸ßÓÚ200 ¡æ¡³£¬²âµÃÓйØʵÑéÊý¾ÝÈçÏ£º£¨±ê×¼×´¿öÏ£©

ʵÑéÐòºÅ

ÑùÆ·µÄÖÊÁ¿/g

NaOHÈÜÒºµÄÌå»ý/mL

°±ÆøµÄÌå»ý/L

¢ñ

7.4

40.00

1.68

¢ò

14.8

40.00

3.36

¢ó

22.2

40.00

1.12

¢ô

37.0

40.00

0.00

£¨1£©ÊµÑé¹ý³ÌÖÐÓйط´Ó¦µÄÀë×Ó·½³ÌʽΪ_________________________________________¡£

£¨2£©ÓÉ¢ñ×éÊý¾ÝÖ±½ÓÍƲ⣺±ê×¼×´¿öÏÂ3.7 g ÑùÆ·½øÐÐͬÑùʵÑéʱ£¬Éú³É°±ÆøµÄÌå»ýΪ__________L¡£

£¨3£©ÊÔ¼ÆËã¸Ã»ìºÏÎïÖУ¨NH4£©2SO4ºÍNH4HSO4µÄÎïÖʵÄÁ¿Ö®±ÈΪ__________________¡£

£¨4£©Óû¼ÆËã¸ÃNaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÓ¦¸ÃÑ¡Ôñ__________×éÊý¾Ý£¬ÓÉ´ËÇóµÃNaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸