ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÓÃÑõ»¯»¹Ô­µÎ¶¨·¨²â¶¨Ä³ÖÖ²ÝËᾧÌ壨H2C2O4¡¤X H2O£©Öнᾧˮ·Ö×ÓÊýµÄʵÑé²½ÖèÈçÏ£º

¢ÙÓ÷ÖÎöÌìƽ³ÆÈ¡²ÝËᾧÌå1.260g£¬½«ÆäÅäÖƳÉ100.00mL´ý²â²ÝËáÈÜÒº£»

¢ÚÓÃÒÆÒº¹ÜÒÆÈ¡25.00mL´ý²â²ÝËáÈÜÒºÓÚ׶ÐÎÆ¿ÖУ¬²¢¼ÓÈëÊÊÁ¿ÁòËáËữ

¢ÛÓÃŨ¶ÈΪ0.1000 mol/LµÄKMnO4±ê×¼ÈÜÒº½øÐе樣¬Èý´Î½á¹ûÈçÏ£º

µÚÒ»´ÎµÎ¶¨

µÚ¶þ´ÎµÎ¶¨

µÚÈý´ÎµÎ¶¨

´ý²âÈÜÒºÌå»ý(mL)

25.00

25.00

25.00

±ê×¼ÈÜÒºÌå»ý(mL)

9.99

10.01

10.00

ÒÑÖª:H2C2O4µÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª90¡£

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©µÎ¶¨Ê±£¬KMnO4±ê×¼ÈÜÒºÓ¦¸Ã×°ÔÚ_____ (Ìî¡°Ëáʽ¡±»ò¡°¼îʽ¡±)µÎ¶¨¹ÜÖС£

£¨2£©H2C2O4ÔÚÈÜÒºÖÐÓëKMnO4ÈÜÒº·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ______________

£¨3£©µ½´ïµÎ¶¨ÖÕµãµÄ±êÖ¾ÊÇ____________¡£

£¨4£©¸ù¾ÝÉÏÊöÊý¾Ý¼ÆËãX=________________¡£

£¨5£©Îó²î·ÖÎö(Ìƫ¸ß¡¢Æ«µÍ¡¢ÎÞÓ°Ïì)£º

¢ÙÈôµÎ¶¨¿ªÊ¼Ê±ÑöÊӵζ¨¹Ü¿Ì¶È£¬µÎ¶¨½áÊøʱ¸©Êӵζ¨¹Ü¿Ì¶È£¬ÔòXÖµ_________£»

¢ÚÈôµÎ¶¨¹ÜˮϴºóÖ±½Ó¼ÓÈëKMnO4±ê×¼ÈÜÒº£¬ÔòXÖµ_______¡£

¡¾´ð°¸¡¿ Ëáʽ 2KMnO4 + 5H2C2O4 + 3H2SO4 K2SO4 + 2MnSO4 + 8H2O + 10CO2¡ü ×îºóÒ»µÎ±ê×¼ÒºµÎÈë׶ÐÎÆ¿ÖУ¬ÈÜҺǡºÃÓÉÎÞÉ«±ä³É×ϺìÉ«£¬ÇÒ30ÃëÄÚ²»ÍÊÉ« 2 Æ«¸ß Æ«µÍ

¡¾½âÎö¡¿¢Å. KMnO4¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬»á¸¯Ê´Ï𽺣¬Òò´ËKMnO4ӦװÔÚËáʽµÎ¶¨¹ÜÖУ¬¹Ê´ð°¸Îª£ºËáʽ£»

¢Æ. KMnO4ÔÚËáÐÔÌõ¼þÏÂÓëH2C2O4·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬MnÔªËصĻ¯ºÏ¼Û´Ó+7¼Û½µµÍµ½+2¼ÛÉú³ÉÁòËáÃÌ£¬CÔªËصĻ¯ºÏ¼Û´Ó+3¼ÛÉý¸ßµ½+4¼ÛÉú³É¶þÑõ»¯Ì¼£¬¸ù¾ÝµÃʧµç×ÓÊغãºÍÔ­×ÓÊغ㣬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£¬2KMnO4 + 5H2C2O4 + 3H2SO4 K2SO4 + 2MnSO4 + 8H2O + 10CO2¡ü£¬¹Ê´ð°¸Îª£º2KMnO4 + 5H2C2O4 + 3H2SO4 K2SO4 + 2MnSO4 + 8H2O + 10CO2¡ü£»

¢Ç.µÎ¶¨µ½´ïÖÕµãʱµÄÏÖÏóΪ£ºµ±×îºóÒ»µÎ±ê×¼ÒºµÎÈë׶ÐÎÆ¿ÖУ¬ÈÜҺǡºÃÓÉÎÞÉ«±ä³É×ϺìÉ«£¬ÇÒ30ÃëÄÚ²»ÍÊÉ«£¬¹Ê´ð°¸Îª£º×îºóÒ»µÎ±ê×¼ÒºµÎÈë׶ÐÎÆ¿ÖУ¬ÈÜҺǡºÃÓÉÎÞÉ«±ä³É×ϺìÉ«£¬ÇÒ30ÃëÄÚ²»ÍÊÉ«£»

¢È.Èý´ÎµÎ¶¨ÏûºÄKMnO4±ê×¼ÒºµÄƽ¾ùÌå»ýΪ£º£¨9.99+10.01+10.00£©mL¡Â3=10.00mL£¬¸ù¾Ý·´Ó¦·½³ÌʽµÃ£ºn(H2C2O4)= ¡Á0.01L¡Á0.1 mol/L¡Á=0.01mol£¬m(H2C2O4)=0.01mol¡Á90g/mol=0.9g£¬Ôòm(H2O)= 1.260g£­0.9g=0.36g£¬n(H2O)= =0.02mol£¬n(H2C2O4): n(H2O)=0.01mol:0.02mol=1:2£¬X=2£¬¹Ê´ð°¸Îª£º2£»

¢É.¢Ù. µÎ¶¨¿ªÊ¼Ê±ÑöÊӵζ¨¹Ü¿Ì¶È£¬µÎ¶¨½áÊøʱ¸©Êӵζ¨¹Ü¿Ì¶È£¬»áÔì³É¶ÁÈ¡±ê×¼ÒºµÄÌå»ýƫС£¬²â¶¨H2C2O4µÄÁ¿Æ«µÍ£¬Ë®µÄÁ¿Æ«¸ß£¬ÔòXÆ«¸ß£¬¹Ê´ð°¸Îª£ºÆ«¸ß£»

¢Ú. µÎ¶¨¹ÜˮϴºóÖ±½Ó¼ÓÈëKMnO4±ê×¼ÈÜÒº£¬Ôì³É±ê×¼ÒºµÄŨ¶ÈƫС£¬ÏûºÄ±ê×¼ÒºµÄÌå»ýÆ«´ó£¬²â¶¨H2C2O4µÄÁ¿Æ«¸ß£¬Ë®µÄÁ¿Æ«µÍ£¬ÔòXÆ«µÍ£¬¹Ê´ð°¸Îª£ºÆ«µÍ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø