ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿îâËáÄÆ£¨Na2MoO4£©ÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤ÔÁÏ¡£Ó÷ϼÓÇâ´ß»¯¼Á£¨º¬ÓÐMoS2ºÍAl2O3¡¢Fe2O3¡¢SiO2µÈ£©ÎªÔÁÏÖÆÈ¡îâËáÄÆ£¬¹¤ÒÕÁ÷³ÌÈçͼËùʾ£º
ÒÑÖª£ºMoO3¡¢A12O3ÔÚ¸ßÎÂÏÂÄܸúNa2CO3·¢Éú·´Ó¦¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)Na2MoO4ÖÐMoÔªËصĻ¯ºÏ¼Û______¡£
(2)·Ï¼ÓÇâ´ß»¯¼Á±ºÉÕÄ¿µÄÊÇÍѳý±íÃæÓÍÖ¬¡¢ÁòµÈ¡£Çë¸ù¾Ý±íÖÐʵÑéÊý¾Ý·ÖÎö£¬·Ï¼ÓÇâ´ß»¯¼ÁÔ¤´¦ÀíζÈӦѡÔñ______¡æ¡£
·Ï´ß»¯¼ÁÔÚ²»Í¬Î¶ÈϵÄÉղУ¨Ê±¼ä£º2h£©
ÎÂ¶È /¡æ | 300 | 350 | 400 | 500 | 600 |
ÉÕÇ° /g | 50.00 | 50.00 | 50.00 | 50.00 | 50.00 |
ÉÕºó /g | 48.09 | 47.48 | 47.19 | 46.55 | 46.52 |
ÉղУ¬ % | 96.2 | 95.0 | 94.4 | 93.1 | 93.0 |
(3)±ºÉÕʱÉú³ÉMoO3µÄ»¯Ñ§·½³ÌʽΪ______£¬µ±Éú³É1mol MoO3תÒƵç×ÓÊýΪ______NA¡£
(4)¼Ó̼ËáÄƼîÐÔ±ºÉÕʱÖ÷Òª·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______¡£
(5)ÓÃ50tº¬MoS2Ϊ80%µÄ·Ï¼ÓÇâ´ß»¯¼Á£¬¾¹ýÖÆÈ¡¡¢·ÖÀë¡¢Ìá´¿£¬µÃµ½30.9t Na2MoO4£¬ÔòNa2MoO4µÄ²úÂÊΪ______¡£
¡¾´ð°¸¡¿+6 500 2MoS2+7O22MoO3+4SO2 14 MoO3+Na2CO3Na2MoO4+CO2¡ü 60%
¡¾½âÎö¡¿
¸ù¾ÝÁ÷³Ì£º¿ÕÆøÖбºÉշϼÓÇâ´ß»¯¼Á£¬MoS2ȼÉÕ·´Ó¦·½³ÌʽΪ2MoS2+7O22MoO3+4SO2£¬¼ÓNa2CO3¼îÐÔ±ºÉÕ£¬ÔÙ¼ÓË®Èܽâ¹ýÂËËùµÃÂËÒº£¬È»ºóÏòÂËÒºÖмÓÈëÏ¡ÁòËáµ÷½ÚpHÖµ£¬¹ýÂË£¬ÈÜÒºÖеÄÈÜÖÊΪNa2MoO4£¬ËùÒÔ¼Ó̼ËáÄƼîÐÔ±ºÉÕʱÖ÷ÒªÊǽ«MoO3ת»¯ÎªÒ×ÈÜÓÚË®µÄNa2MoO4£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪMoO3+Na2CO3Na2MoO4+CO2¡ü£¬½«ÈÜÒºÕô·¢Å¨Ëõ¡¢ÀäÈ´½á¾§£¬µÃµ½Na2MoO42H2O¾§Ì壬ÔÙ·Ö½âµÃµ½Na2MoO4£¬¾Ý´Ë·ÖÎö×÷´ð¡£
(1)¸ù¾ÝNa2MoO4Öл¯ºÏ¼Û´úÊýºÍΪ0£¬MoÔªËصĻ¯ºÏ¼ÛΪ+£¨2¡Á4-1¡Á2£©=+6¼Û£»
¹Ê´ð°¸Îª£º+6£»
(2)ÓɷϼÓÇâ´ß»¯¼Á±ºÉÕÄ¿µÄÊÇÍѳý±íÃæÓÍÖ¬¡¢ÁòµÈ£¬½áºÏ±íÖÐʵÑéÊý¾Ý¿ÉÖª£¬500¡æºóÉղеİٷֺ¬Á¿¼¸ºõ²»±ä£¬ËùÒԷϼÓÇâ´ß»¯¼ÁÔ¤´¦ÀíζÈӦѡÔñ500¡æ£¬¹Ê´ð°¸Îª£º500£»
(3)¿ÕÆøÖбºÉշϼÓÇâ´ß»¯¼Á£¬MoS2ȼÉÕ·´Ó¦Éú³ÉMoO3µÄ»¯Ñ§·½³ÌʽΪ2MoS2+7O22MoO3+4SO2£¬·´Ó¦ÖÐMoÔªËØ»¯ºÏ¼ÛÓÉ+4¼ÛÉý¸ßµ½+6£¬SÔªËØ»¯ºÏ¼ÛÓÉ-2Éý¸ßµ½+4£¬ÔòÓÉ·½³Ìʽ¿É֪תÒƵç×ÓÊýΪ2¡Á£¨6-4+2¡Á6£©=28e-£¬ËùÒÔµ±Éú³É1mol MoO3תÒƵç×ÓÊýΪ14NA£»
¹Ê´ð°¸Îª£º2MoS2+7O22MoO3+4SO2£»14£»
(4)¼Ó̼ËáÄƼîÐÔ±ºÉÕʱÖ÷ÒªÊǽ«MoO3ת»¯ÎªÒ×ÈÜÓÚË®µÄNa2MoO4£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪMoO3+Na2CO3Na2MoO4+CO2¡ü£»
¹Ê´ð°¸Îª£ºMoO3+Na2CONa2MoO4+CO2¡ü£»
(5)ÓÃ50tº¬MoS2Ϊ80%µÄ·Ï¼ÓÇâ´ß»¯¼Á£¬Ôòº¬MoS2µÄÎïÖʵÄÁ¿Îª=2.5¡Á105mol£¬¸ù¾ÝMoÔ×ÓÊغãn£¨Mo£©=n£¨MoS2£©=n£¨Na2MoO4£©£¬ËùÒÔNa2MoO4µÄ²úÂÊ=¡Á100%=60%£¬¹Ê´ð°¸Îª£º60%¡£
¡¾ÌâÄ¿¡¿Ó뻯ѧƽºâÀàËÆ£¬µçÀëƽºâµÄƽºâ³£Êý£¬½Ð×öµçÀë³£Êý£¨ÓÃK±íʾ£©¡£Ï±íÊÇijζÈϼ¸ÖÖ³£¼ûÈõËáµÄµçÀëƽºâ³£Êý£º
Ëá | µçÀë·½³Ìʽ | µçÀëƽºâ³£ÊýK |
CH3COOH | CH3COOHCH3COO-+H+ | 2¡Á10-5 |
HClO | HClOClO-+H+ | 3.0¡Á10-8 |
H2CO3 | H2CO3H++HCO3- HCO3-H++CO32- | K1=4.4¡Á10-7 K2=5.6¡Á10-11 |
H3PO4 | H3PO4H++H2PO4- H2PO4-H++HPO42- HPO42-H++PO43- | K1=7.1¡Á10-3 K2=6.3¡Á10-8 K3=4.2¡Á10-13 |
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Èô°ÑCH3COOH¡¢HClO¡¢H2CO3¡¢HCO3-¡¢H3PO4¡¢H2PO4-¡¢HPO42-¶¼¿´×÷ÊÇËᣬÔòËüÃÇËáÐÔ×îÇ¿µÄÊÇ___________£¨Ìѧʽ£©¡£
£¨2£©ÏòNaClOÈÜÒºÖÐͨÈëÉÙÁ¿µÄ¶þÑõ»¯Ì¼£¬·¢ÉúµÄÀë×Ó·½³ÌʽΪ________¡£
£¨3£©¸ÃζÈÏÂ0.1mol/LµÄCH3COOHÈÜÒº¼ÓˮϡÊ͹ý³Ì£¬ÏÂÁбí´ïʽµÄÊý¾ÝÒ»¶¨Ôö´óµÄÊÇ__¡£
A c£¨H+£© B c£¨H+£©c£¨CH3COO
£¨4£©È¡µÈÌå»ýµÄpH¾ùΪaµÄ´×ËáºÍ´ÎÂÈËáÁ½ÈÜÒº£¬·Ö±ðÓõÈŨ¶ÈµÄNaOHÏ¡ÈÜҺǡºÃÖкͣ¬ÏûºÄµÄNaOHÈÜÒºµÄÌå»ý·Ö±ðΪV1£¬V2£¬Ôò´óС¹ØϵΪ£ºV1_________V2£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©¡£
£¨5£©ÏÂÁÐËÄÖÖÀë×Ó½áºÏH+ÄÜÁ¦×îÇ¿µÄÊÇ________¡£
A HCO3- B CO32- C ClO- D CH3COO-
£¨6£©µÈÎïÖʵÄÁ¿µÄ¿ÁÐÔÄÆ·Ö±ðÓÃpHΪ2ºÍ3µÄ´×ËáÈÜÒºÖкͣ¬ÉèÏûºÄ´×ËáÈÜÒºµÄÌå»ýÒÀ´ÎΪVa¡¢Vb£¬ÔòÁ½ÕߵĹØϵÕýÈ·µÄÊÇ_________¡£
A Va£¾10Vb B Va£¼10Vb C Vb£¼10Va D Vb£¾10Va
£¨7£©ÒÑÖª100¡æʱ£¬Ë®µÄÀë×Ó»ý³£ÊýKw=1.0¡Á10-12£¬pH=3µÄCH3COOHºÍpH=9µÄNaOHÈÜÒºµÈÌå»ý»ìºÏ£¬»ìºÏÈÜÒº³Ê_____________ÐÔ£»
£¨8£©µÈŨ¶ÈµÄ¢Ù£¨NH4£©2SO4¡¢¢ÚNH4HSO4¡¢¢ÛNH4HCO3¡¢¢ÜNH4Cl¡¢¢ÝNH3H2OÈÜÒºÖУ¬NH4+Ũ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ£º_________¡£
£¨9£©¼ÆËã¸ÃζÈÏÂCH3COONaµÄË®½âƽºâ³£ÊýKh_________¡£
£¨10£©ÎïÖʵÄÁ¿Å¨¶ÈÏàͬµÄËÄÖÖÈÜÒº£ºa CH3COONa¡¢b NaHCO3¡¢c NaClO¡¢d Na2CO3
ËÄÖÖÈÜÒºµÄpHÓÉСµ½´óÅÅÁеÄ˳ÐòÊÇ____________£¨ÓñàºÅÌîд£©¡£