ÌâÄ¿ÄÚÈÝ
£¨16·Ö£©ÒÑÖªFeSO4ÔÚ²»Í¬Ìõ¼þÏ·ֽâµÃµ½µÄ²úÎﲻͬ£¬¿ÉÄÜÊÇFeOºÍSO3£¬Ò²¿ÉÄÜÊÇFe2O3¡¢SO3ºÍSO2¡£Ä³Ñо¿Ð¡×é̽¾¿Ôھƾ«ÅçµÆ¼ÓÈÈÌõ¼þÏÂFeSO4·Ö½âµÄÆøÌå²úÎï¡£ÒÑÖªSO3µÄÈÛµãÊÇ16.8¡æ£¬·ÐµãÊÇ44.8¡æ¡£
£¨1£©×°ÖâòµÄÊÔ¹ÜÖв»×°ÈκÎÊÔ¼Á£¬Æä×÷ÓÃÊÇ______________________
ÊԹܽþÅÝÔÚ50¡æµÄˮԡÖУ¬Ä¿µÄÊÇ______________________________
£¨2£©×°ÖâóºÍ×°ÖâôµÄ×÷ÓÃÊÇ̽¾¿±¾ÊµÑéÆøÌå²úÎï³É·Ö¡£ÇëÍê³ÉʵÑéÉè¼Æ£¬Ìîд¼ìÑéÊÔ¼Á¡¢Ô¤ÆÚÏÖÏóÓë½áÂÛ¡£
ÏÞÑ¡ÊÔ¼Á£º3 mol¡¤L£1 H2SO4¡¢6 mol¡¤L£1 NaOH¡¢0.5 mol¡¤L£1 BaCl2¡¢0.5 mol¡¤L£1 Ba(NO3)2¡¢
0.01 mol¡¤L£1ËáÐÔKMnO4ÈÜÒº¡¢0.01 mol¡¤L£1äåË®¡£
£¨3£©×°ÖâõµÄ×÷ÓÃÊÇ·ÀֹβÆøÎÛȾ»·¾³£¬ÉÕ±ÖÐÓ¦¼ÓÈëµÄÊÔ¼ÁÊÇ ¡£
£¨1£©×°ÖâòµÄÊÔ¹ÜÖв»×°ÈκÎÊÔ¼Á£¬Æä×÷ÓÃÊÇ______________________
ÊԹܽþÅÝÔÚ50¡æµÄˮԡÖУ¬Ä¿µÄÊÇ______________________________
£¨2£©×°ÖâóºÍ×°ÖâôµÄ×÷ÓÃÊÇ̽¾¿±¾ÊµÑéÆøÌå²úÎï³É·Ö¡£ÇëÍê³ÉʵÑéÉè¼Æ£¬Ìîд¼ìÑéÊÔ¼Á¡¢Ô¤ÆÚÏÖÏóÓë½áÂÛ¡£
ÏÞÑ¡ÊÔ¼Á£º3 mol¡¤L£1 H2SO4¡¢6 mol¡¤L£1 NaOH¡¢0.5 mol¡¤L£1 BaCl2¡¢0.5 mol¡¤L£1 Ba(NO3)2¡¢
0.01 mol¡¤L£1ËáÐÔKMnO4ÈÜÒº¡¢0.01 mol¡¤L£1äåË®¡£
¼ìÑéÊÔ¼Á | Ô¤ÆÚÏÖÏóºÍ½áÂÛ |
×°ÖâóµÄÊÔ¹ÜÖмÓÈë__________ ___¡£ | ²úÉú´óÁ¿°×É«³Áµí£¬Ö¤Ã÷ÆøÌå²úÎïÖк¬ÓÐSO3¡£ |
×°ÖâôµÄÊÔ¹ÜÖмÓÈë_______ _________¡£ | ______________________________ ______________________________ ______________________________ ______________________________ |
£¨16·Ö£©
£¨1£©·ÀÖ¹ÈÜÒºµ¹ÎüÈë×°Öâñ£¨2·Ö£© ·ÀÖ¹SO3Òº»¯»òÄý¹Ì£¨2·Ö£©
£¨2£©£¨10·Ö£©
»ò£º
£¨3£©NaOHÈÜÒº£¨2·Ö£©
£¨1£©·ÀÖ¹ÈÜÒºµ¹ÎüÈë×°Öâñ£¨2·Ö£© ·ÀÖ¹SO3Òº»¯»òÄý¹Ì£¨2·Ö£©
£¨2£©£¨10·Ö£©
¼ìÑéÊÔ¼Á | Ô¤ÆÚÏÖÏóºÍ½áÂÛ |
×°ÖâóµÄÊÔ¹ÜÖÐ×°ÓÐBaCl2ÈÜÒº¡££¨3·Ö£© | ²úÉú´óÁ¿°×É«³Áµí£¬Ö¤Ã÷ÆøÌå²úÎïÖк¬ÓÐSO3¡£ |
×°ÖâôµÄÊÔ¹ÜÖÐ×°ÓÐËáÐÔKMnO4ÈÜÒº¡££¨3·Ö£© | ÈôÈÜÒº×ϺìÉ«ÍÊÈ¥£¬Ö¤Ã÷ÆøÌå²úÎïÖк¬ÓÐSO2£»£¨2·Ö£© ÈôÈÜÒº×ϺìÉ«ÎÞÃ÷ÏԱ仯£¬Ö¤Ã÷ÆøÌå²úÎïÖв»º¬SO2£»£¨2·Ö£© |
¼ìÑéÊÔ¼Á | Ô¤ÆÚÏÖÏóºÍ½áÂÛ |
×°ÖâóµÄÊÔ¹ÜÖÐ×°ÓÐBaCl2ÈÜÒº¡££¨3·Ö£© | ²úÉú´óÁ¿°×É«³Áµí£¬Ö¤Ã÷ÆøÌå²úÎïÖк¬ÓÐSO3¡£ |
×°ÖâôµÄÊÔ¹ÜÖÐ×°ÓÐäåË®¡££¨3·Ö£© | ÈôäåË®³ÈÉ«ÍÊÈ¥£¬Ö¤Ã÷ÆøÌå²úÎïÖк¬ÓÐSO2£»£¨2·Ö£© ÈôäåË®³ÈÉ«ÎÞÃ÷ÏԱ仯£¬Ö¤Ã÷ÆøÌå²úÎïÖв»º¬SO2£»£¨2·Ö£© |
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿