Åðþ¿óÊôÓÚÅðËáÑΣ¬¿ÉÓÃÀ´ÖƱ¸ÅðËᣨH3BO3£©ºÍMgO£¬·½·¨ÈçÏ£ºÅðþ¿ó·ÛÓë(NH4)2SO4ÈÜÒº»ìºÏºó¼ÓÈÈ£¬·´Ó¦Éú³ÉH3BO3¾§ÌåºÍMgSO4ÈÜÒº£¬Í¬Ê±·Å³öNH3£»ÔÙÏòMgSO4ÈÜÒºÖÐͨÈëNH3ÓëCO2£¬µÃµ½MgCO3³ÁµíºÍÂËÒº£¬³Áµí¾­Ï´µÓ¡¢ìÑÉÕºóµÃMgO£¬ÂËÒºÔòÑ­»·Ê¹Ó᣻شðÏÂÁÐÎÊÌ⣺

£¨1£©Óë¹èËáÑÎÀàËÆ£¬ÅðËáÑνṹҲ±È½Ï¸´ÔÓ£¬ÈçÓ²Åð¸Æʯ»¯Ñ§Ê½ÎªCa2B6O11¡¤5H2O£¬½«Æä¸ÄдΪÑõ»¯ÎïµÄÐÎʽ    ¡£

£¨2£©ÉÏÊöÖƱ¸¹ý³ÌÖУ¬¼ìÑé³ÁµíÏ´µÓÊÇ·ñÍêÈ«µÄ·½·¨ÊÇ    ¡£

£¨3£©Ð´³öMgSO4ÈÜÒºÖÐͨÈëNH3ÓëCO2·´Ó¦µÄ»¯Ñ§·½³Ìʽ    ¡£

£¨4£©Èô׼ȷ³ÆÈ¡1.68 gÅðþ¿ó£¬ÍêÈ«·´Ó¦ºóµÃH3BO3¾§Ìå1.24 g£¬MgO 0.8 g£¬¼ÆËã¸ÃÅðËáÑεÄ×é³É¡££¨Ð´³ö¼ÆËã¹ý³Ì£©   

 

¡¾´ð°¸¡¿

 

£¨1£©2CaO¡¤3B2O3¡¤5H2O£¨2·Ö£©

£¨2£©È¡ÉÙÐí×îºóÒ»´ÎÏ´µÓÂËÒº£¬µÎÈë1¡«2µÎBa(NO3)2ÈÜÒº£¬Èô²»³öÏÖ°×É«»ë×Ç£¬±íʾÒÑÏ´µÓÍêÈ«¡££¨2·Ö£©

£¨3£©MgSO4+CO2+2NH3+H2O=MgCO3+(NH4)2SO4£¨2·Ö£©

£¨4£©n (B2O3)=0.01mol£¨1·Ö£©

n (MgO)=0.02mol £¨1·Ö£©

n (H2O) = (1.68 g -0.7 g -0.8 g) /18 g¡¤mol-1= 0.01mol £¨2·Ö£©

¹ÊÆä×é³ÉΪ£ºMg2B2O5¡¤H2O»ò2MgO¡¤B2O3¡¤H2O¡££¨2·Ö£©

£¨ËµÃ÷£ºÆäËüºÏÀí½â·¨Òà¿É£¬B¡¢MgÎïÖʵÄÁ¿¸÷1·Ö£¬Ñõ¡¢Ë®µÄÎïÖʵÄÁ¿¸÷1·Ö£¬×é³É2·Ö£©

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£º£¨1£©°´ÕÕ»¯Ñ§Ê½ÖÐÔªËصÄ˳Ðò£¬ÒÀ´Îд³öÑõ»¯ÎïµÄÐÎʽ£¬×¢ÒâÔªËصĻ¯ºÏ¼ÛºÍÔ­×ÓÊýÄ¿µÄ¹Øϵ£»£¨2£©¼ìÑé³ÁµíÏ´µÓÊÇ·ñÍêÈ«µÄ·½·¨Òª×¢ÒâÈ¡×îºÃÒ»´ÎµÄÏ´µÓÒº£¬¼ÓÈëÊÔ¼Á£¨Ñ¡Ôñ¼ìÑéÔÓÖÊÀë×ÓµÄ×îÃ÷ÏÔ¡¢ÁéÃôµÄÊÔ¼Á£©£¬»¹ÒªËµÃ÷ÏÖÏó½áÂÛµÈÄÚÈÝ£»£¨4£©¸ÃÌâµÄµÚ£¨1£©ÎÊÒÑÌá³öÑõ»¯ÎïµÄÐÎʽ£¬¹Ê¸ÃÌâµÄ¼ÆËãÓ¦¼ÆËãþ¡¢¹èÏàÓ¦Ñõ»¯ÎïµÄÁ¿£¬ÔÙÈ·¶¨ÊÇ·ñº¬ÓÐË®¡£

n (B2O3)=0.01mol

n (MgO)=0.02mol

n (H2O) = (1.68 g -0.7 g -0.8 g) /18 g¡¤mol-1= 0.01mol

¹ÊÆä×é³ÉΪ£ºMg2B2O5¡¤H2O»ò2MgO¡¤B2O3¡¤H2O¡£

¿¼µã£º¿¼²éÎïÖʵÄÖƱ¸Öз´Ó¦·½³ÌʽµÄÊéд¡¢ÊµÑé²Ù×÷¡¢»¯Ñ§Ê½µÄÈ·¶¨µÈÓйØÎÊÌâ¡£

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Åðþ¿óÊôÓÚÅðËáÑΣ¬¿ÉÓÃÀ´ÖƱ¸ÅðËᣨH3BO3£©ºÍMgO£¬·½·¨ÈçÏ£ºÅðþ¿ó·ÛÓ루NH4£©2SO4ÈÜÒº»ìºÏºó¼ÓÈÈ£¬·´Ó¦Éú³ÉH3BO3¾§ÌåºÍMgSO4ÈÜÒº£¬Í¬Ê±·Å³öNH3£»ÔÙÏòMgSO4ÈÜÒºÖÐͨÈëNH3ÓëCO2£¬µÃµ½MgCO3³ÁµíºÍÂËÒº£¬³Áµí¾­Ï´µÓ¡¢ìÑÉÕºóµÃMgO£¬ÂËÒºÔòÑ­»·Ê¹Ó㮻شðÏÂÁÐÎÊÌ⣺
£¨1£©Óë¹èËáÑÎÀàËÆ£¬ÅðËáÑνṹҲ±È½Ï¸´ÔÓ£¬ÈçÓ²Åð¸Æʯ»¯Ñ§Ê½ÎªCa2B6O11?5H2O£¬½«Æä¸ÄдΪÑõ»¯ÎïµÄÐÎʽΪ
2CaO?3B2O3?5H2O
2CaO?3B2O3?5H2O
£®
£¨2£©ÉÏÊöÖƱ¸¹ý³ÌÖУ¬¼ìÑé³ÁµíÏ´µÓÊÇ·ñÍêÈ«µÄ·½·¨ÊÇ
È¡ÉÙÐí×îºóÒ»´ÎÏ´µÓÂËÒº£¬µÎÈë1¡«2µÎBa£¨NO3£©2ÈÜÒº£¬Èô²»³öÏÖ°×É«»ë×Ç£¬±íʾÒÑÏ´µÓÍêÈ«
È¡ÉÙÐí×îºóÒ»´ÎÏ´µÓÂËÒº£¬µÎÈë1¡«2µÎBa£¨NO3£©2ÈÜÒº£¬Èô²»³öÏÖ°×É«»ë×Ç£¬±íʾÒÑÏ´µÓÍêÈ«
£®
£¨3£©Ð´³öMgSO4ÈÜÒºÖÐͨÈëNH3ÓëCO2·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
MgSO4+CO2+2NH3+H2O¨TMgCO3+£¨NH4£©2SO4
MgSO4+CO2+2NH3+H2O¨TMgCO3+£¨NH4£©2SO4
£®
£¨4£©Èô׼ȷ³ÆÈ¡1.68gÅðþ¿ó£¬ÍêÈ«·´Ó¦ºóµÃH3BO3¾§Ìå1.24g£¬MgO 0.8g£¬¼ÆËã¸ÃÅðËáÑεÄ×é³É£®£¨Ð´³ö¼ÆËã¹ý³Ì£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

A¡¢B¡¢C¡¢D¡¢EÎåÖÖÌþ·Ö×ÓÖоùº¬ÓÐ8¸öÇâÔ­×Ó£¬ÆäÖÐA¡¢B³£ÎÂϳÊÆø̬£¬C¡¢D¡¢E³ÊҺ̬£®
£¨1£©AÊÇ·ûºÏÉÏÊöÌõ¼þÖÐÏà¶Ô·Ö×ÓÖÊÁ¿×îСµÄ£¬ÔòAµÄ·Ö×ÓʽΪ
C3H8
C3H8
£»BÊôÓÚÁ´×´µ¥Ï©Ìþ£¬ÇÒÓëHBr¼Ó³ÉµÄ²úÎïÖ»ÓÐÒ»ÖÖ£¬ÊÔд³öBµÄ½á¹¹¼òʽ
CH3CH=CHCH3
CH3CH=CHCH3
£®
£¨2£©CÊôÓÚ·¼ÏãÌþ£¬¿ÉÓÃÀ´ÖÆÁÒÐÔÕ¨Ò©£¬ÊÔд³öCÖÆÕ¨Ò©µÄ»¯Ñ§·½³Ìʽ
£®
£¨3£©ÒÑÖª£º

D´æÔÚÈçÏÂת»¯¹Øϵ£ºD
¢ÙO3
¢ÚZn£¬H2O
OHC-CH2-CHO£¨Î¨Ò»²úÎ£¬ÊÔд³öDµÄ½á¹¹¼òʽ
£®
£¨4£©E´æÔÚÓÚú½¹ÓÍÖУ¬EÖÐËùÓÐÔ­×Ó¾ù´¦ÓÚͬһƽÃæÄÚ£¬1molE×î¶àÄܺÍ5molH2¼Ó³É£¬ÊÔд³öEµÄ½á¹¹¼òʽ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

A£®B£®C£®D£®EÎåÖÖÌþ·Ö×ÓÖоùº¬ÓÐ8¸öÇâÔ­×Ó£¬ÆäÖÐA£®B³£ÎÂϳÊÆø̬£¬C£®D£®E³ÊҺ̬£®
£¨1£©AÊÇ·ûºÏÉÏÊöÌõ¼þÖÐÏà¶Ô·Ö×ÓÖÊÁ¿×îСµÄ£¬ÔòAµÄ·Ö×ÓʽΪ
C3H8
C3H8
£»BÊôÓÚÁ´×´µ¥Ï©Ìþ£¬ÇÒÓëHBr¼Ó³ÉµÄ²úÎïÖ»ÓÐÒ»ÖÖ£¬ÊÔд³öBµÄ½á¹¹¼òʽ
CH3CH=CHCH3
CH3CH=CHCH3

£¨2£©CÊôÓÚ·¼ÏãÌþ£¬¿ÉÓÃÀ´ÖÆÁÒÐÔÕ¨Ò©£¬ÊÔд³öCÖÆÈ¡ÁÒÐÔÕ¨Ò©µÄ·½³Ìʽ

£¨3£©ÒÑÖªEµÄ½á¹¹¼òʽΪ ´æÔÚÓÚú½¹ÓÍÖУ¬ËùÓÐÔ­×Ó¾ù´¦ÓÚͬһƽÃæÄÚ£¬Ôò1mol E ×î¶àÄܺÍ
5
5
 molµÄH2¼Ó³É£¬Ð´³öEµÄÒ»ÂÈ´úÎïµÄËùÓеĽṹ¼òʽ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

Åðþ¿óÊôÓÚÅðËáÑΣ¬¿ÉÓÃÀ´ÖƱ¸ÅðËᣨH3BO3£©ºÍMgO£¬·½·¨ÈçÏ£ºÅðþ¿ó·ÛÓ루NH4£©2SO4ÈÜÒº»ìºÏºó¼ÓÈÈ£¬·´Ó¦Éú³ÉH3BO3¾§ÌåºÍMgSO4ÈÜÒº£¬Í¬Ê±·Å³öNH3£»ÔÙÏòMgSO4ÈÜÒºÖÐͨÈëNH3ÓëCO2£¬µÃµ½MgCO3³ÁµíºÍÂËÒº£¬³Áµí¾­Ï´µÓ¡¢ìÑÉÕºóµÃMgO£¬ÂËÒºÔòÑ­»·Ê¹Ó㮻شðÏÂÁÐÎÊÌ⣺
£¨1£©Óë¹èËáÑÎÀàËÆ£¬ÅðËáÑνṹҲ±È½Ï¸´ÔÓ£¬ÈçÓ²Åð¸Æʯ»¯Ñ§Ê½ÎªCa2B6O11?5H2O£¬½«Æä¸ÄдΪÑõ»¯ÎïµÄÐÎʽΪ______£®
£¨2£©ÉÏÊöÖƱ¸¹ý³ÌÖУ¬¼ìÑé³ÁµíÏ´µÓÊÇ·ñÍêÈ«µÄ·½·¨ÊÇ______£®
£¨3£©Ð´³öMgSO4ÈÜÒºÖÐͨÈëNH3ÓëCO2·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º______£®
£¨4£©Èô׼ȷ³ÆÈ¡1.68gÅðþ¿ó£¬ÍêÈ«·´Ó¦ºóµÃH3BO3¾§Ìå1.24g£¬MgO 0.8g£¬¼ÆËã¸ÃÅðËáÑεÄ×é³É£®£¨Ð´³ö¼ÆËã¹ý³Ì£©

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸