ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿ÔËÓÃÒÑѧ֪ʶÍê³ÉÏÂÁмÆË㣺
£¨1£©17gNH3¹²ÓÐ______molÔ×Ó£¬0.1molH2S¹²ÓÐ_____¸öÇâÔ×Ó£»Í¬ÖÊÁ¿µÄNH3ºÍH2SÖзÖ×Ó¸öÊý±ÈΪ_______¡£
£¨2£©ÔÚ±ê×¼×´¿öÏ£¬1.7g°±ÆøËùÕ¼µÄÌå»ýΪ_____L£¬ËüÓë±ê×¼×´¿öÏÂ_____LµÄÁò»¯Ç⣨H2S£©º¬ÓÐÏàͬÊýÄ¿µÄÇâÔ×Ó¡£
£¨3£©ÏÖÓÐmgijÆøÌ壬ËüÓÉË«Ô×Ó·Ö×Ó¹¹³É£¬ËüµÄĦ¶ûÖÊÁ¿ÎªMgmol-1¡£Èô°¢·ü¼ÓµÂÂÞ³£ÊýÓÃNA±íʾ£¬Ôò¸ÃÆøÌåµÄÎïÖʵÄÁ¿Îª_____mol£¬¸ÃÆøÌåËùº¬Ô×Ó×ÜÊýΪ______¸ö£¬ÔÚ±ê×¼×´¿öϸÃÆøÌåµÄÌå»ýΪ________L¡£Èô¸ÃÆøÌå²»ÓëË®·´Ó¦£¬½«ÆäÈÜÓÚ1LË®ÖУ¬ËùµÃÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪ_________¡£
¡¾´ð°¸¡¿4 0.2NA 2:1 2.24 3.36 ¡Á100%
¡¾½âÎö¡¿
£¨1£©17gNH3µÄÎïÖʵÄÁ¿ÊÇ =1mol£¬1¸öNH3·Ö×ÓÖÐÓÐ4¸öÔ×Ó£¬ËùÒÔ1mol°±Æø¹²ÓÐ4molÔ×Ó£¬1¸öH2SÖÐÓÐ2¸öHÔ×Ó£¬0.1molH2S¹²ÓÐ0.2NA¸öÇâÔ×Ó£»Í¬ÖÊÁ¿µÄNH3ºÍH2S£¬ÉèÖÊÁ¿¶¼ÊÇmg£¬ÔòNH3ºÍH2SµÄÎïÖʵÄÁ¿·Ö±ðÊÇ ¡¢£¬·Ö×Ó¸öÊý±ÈµÈÓÚÎïÖʵÄÁ¿±È£¬·Ö×ÓÊý±ÈΪ£º=2:1¡£
£¨2£©1.7gNH3µÄÎïÖʵÄÁ¿ÊÇ =0.1mol£¬ÔÚ±ê×¼×´¿öÏÂËùÕ¼µÄÌå»ýΪ0.1mol¡Á22.4L/mol=2.24L£»0.1mol°±ÆøÖк¬ÓÐÇâÔ×ÓµÄÎïÖʵÄÁ¿ÊÇ0.3mol£¬º¬ÓÐ0.3molÇâÔ×ÓµÄH2SµÄÎïÖʵÄÁ¿ÊÇ0.15mol£¬ÔÚ±ê×¼×´¿öϵÄÌå»ýÊÇ0.15mol¡Á22.4L/mol=3.36L¡£
£¨3£©mgijÆøÌ壬ËüµÄĦ¶ûÖÊÁ¿ÎªMgmol-1£¬Ôò¸ÃÆøÌåµÄÎïÖʵÄÁ¿Îªmol£¬¸ÃÆøÌåΪ˫Ô×Ó·Ö×Ó£¬Ëùº¬Ô×ÓµÄÎïÖʵÄÁ¿ÊÇmol£¬Ô×ÓÊýΪ¸ö£¬ÔÚ±ê×¼×´¿öϸÃÆøÌåµÄÌå»ýΪmol¡Á22.4L/mol= L¡£Èô¸ÃÆøÌå²»ÓëË®·´Ó¦£¬½«ÆäÈÜÓÚ1LË®ÖУ¬ËùµÃÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿ÊÇmg¡¢ÈÜÒºÖÊÁ¿ÊÇ(1000+m)g£¬ÈÜÖʵÄÖÊÁ¿·ÖÊýΪ¡Á100%¡£
¡¾ÌâÄ¿¡¿ÒÑÖª£ºCO(g)£«H2O(g)CO2(g)£«H2(g)¡¡¦¤H£½£41 kJ¡¤mol£1¡£ÏàͬζÈÏ£¬ÔÚÈÝ»ýÏàͬµÄÁ½¸öºãÎÂÃܱÕÈÝÆ÷ÖУ¬¼ÓÈëÒ»¶¨Á¿µÄ·´Ó¦Îï·¢Éú·´Ó¦¡£Ïà¹ØÊý¾ÝÈçÏ£º
ÈÝÆ÷±àºÅ | Æðʼʱ¸÷ÎïÖÊÎïÖʵÄÁ¿/mol | ´ïƽºâ¹ý³ÌÌåϵÄÜÁ¿µÄ±ä»¯ | |||
CO | H2O | CO2 | H2 | ||
¢Ù | 1 | 4 | 0 | 0 | ·Å³öÈÈÁ¿£º32.8 kJ |
¢Ú | 0 | 0 | 1 | 4 | ÈÈÁ¿±ä»¯£ºQ kJ |
ÏÂÁÐ˵·¨ÖУ¬²»ÕýÈ·µÄÊÇ(¡¡¡¡)
A. ÈÝÆ÷¢ÙÖз´Ó¦´ïƽºâʱ£¬COµÄת»¯ÂÊΪ80%
B. ÈÝÆ÷¢ÙÖÐCOµÄת»¯ÂʵÈÓÚÈÝÆ÷¢ÚÖÐCO2µÄת»¯ÂÊ
C. ƽºâʱ£¬Á½ÈÝÆ÷ÖÐCO2µÄŨ¶ÈÏàµÈ
D. ÈÝÆ÷¢ÙʱCOµÄ·´Ó¦ËÙÂʵÈÓÚH2OµÄ·´Ó¦ËÙÂÊ