ÌâÄ¿ÄÚÈÝ

18£®Èô¼×¡¢ÒÒ¡¢±ûÈýλͬѧÓù涨µÄÒ©Æ·ÖÆ±¸Al£¨OH£©3£¬¹æ¶¨±ØÓõÄÒ©Æ·ÈçÏ£º350g 70% H2SO4ÈÜÒº£¬NaOH¹ÌÌå240g£¬×ãÁ¿ÂÁм£¬Ë®£¨²»ÄÜÓÃÆäËûÒ©Æ·£©£®
¼×¡¢ÒÒ¡¢±ûÓø÷×ÔÉè¼ÆµÄ·½°¸ÖƵÃAl£¨OH£©3µÄÖÊÁ¿·Ö±ðÊÇW1£¬W2£¬W3£®ÈýÖÖʵÑé·½°¸ÈçÏ£º
¼×£ºÂÁ¡ú¼ÓNaOHÈÜÒº¡ú¼ÓH2SO4ÈÜÒº¡úW1g Al£¨OH£©3
ÒÒ£ºÂÁ¡ú¼ÓH2SO4ÈÜÒº¡ú¼ÓNaOHÈÜÒº¡úW2g Al£¨OH£©3
±û£º$\left.\begin{array}{l}{{ÂÁ¡ú¼ÓH}_{2}S{O}_{4}ÈÜÒº}\\{ÂÁ¡ú¼ÓNaOHÈÜÒº}\end{array}\right\}$¡úW3g Al£¨OH£©3£®
ÊԻشð£º
£¨1£©´Ó³ä·ÖÀûÓÃÔ­ÁÏ¡¢½µµÍ³É±¾ºÍÌá¸ß²úÂʵÈÒòËØ·ÖÎö£¬ÊµÑé·½°¸×îºÏÀíµÄÊDZû£»
£¨2£©ÈýλѧÉúÖÆµÃµÄAl£¨OH£©3£¬W1£¬W2£¬W3ÆäÖÊÁ¿ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇW3£¾W1£¾W2£»
£¨3£©ÖƵÃAl£¨OH£©3µÄ×î´óÖÊÁ¿ÊÇ520g£®

·ÖÎö ÁòËáµÄÎïÖʵÄÁ¿Îª$\frac{350g¡Á70%}{98g/mol}$=2.5mol£¬NaOHµÄÎïÖʵÄÁ¿$\frac{240g}{40g/mol}$=6mol£¬
¼×ÖйØÏµÊ½Îª£ºAl¡«OH-¡«AlO2-¡«H+¡«Al£¨OH£©3¡ý£¬ÁòËá¹ýÁ¿£¬´æÔÚ¹ØÏµÊ½£ºAl£¨OH£©3¡«3H+¡«Al3+£¬
ÒÒÖйØÏµÊ½Îª£ºAl¡«3H+¡«Al3+¡«3OH-¡«Al£¨OH£©3¡ý£¬ÇâÑõ»¯ÄƹýÁ¿£¬´æÔÚ¹ØÏµÊ½£ºAl£¨OH£©3¡ý¡«OH-¡«AlO2-£¬
±ûÖйØÏµÊ½Îª£ºAl¡«3H+¡«Al3+£¬Al¡«OH-¡«AlO2-£¬3AlO2-¡«Al3+¡«4Al£¨OH£©3¡ý£¬
½áºÏ¹ØÏµÊ½·ÖÎö¼ÆËã½â´ð£®

½â´ð ½â£º£¨1£©¼×ÖйØÏµÊ½Îª£ºAl¡«OH-¡«AlO2-¡«H+¡«Al£¨OH£©3¡ý£¬ÖƱ¸2molAl£¨OH£©3£¬ÏûºÄ2molOH-¡¢2molH+£»
ÒÒÖйØÏµÊ½Îª£ºAl¡«3H+¡«Al3+¡«3OH-¡«Al£¨OH£©3¡ý£¬ÖƱ¸2molAl£¨OH£©3£¬ÏûºÄ6molOH-¡¢6molH+£»
±ûÖйØÏµÊ½Îª£ºAl¡«3H+¡«Al3+£¬Al¡«OH-¡«AlO2-£¬3AlO2-¡«Al3+¡«4Al£¨OH£©3¡ý£¬ÖƱ¸2molAl£¨OH£©3£¬ÏûºÄ1.5molOH-¡¢1.5molH+£¬
ÖÆ±¸ÏàͬÎïÖʵÄÁ¿µÄÖÆ±¸Al£¨OH£©3£¬±ûÏûºÄËá¡¢¼î×îÉÙ£¬¹ÊʵÑé·½°¸×îºÏÀíµÄÊÇ£º±û£¬¹Ê´ð°¸Îª£º±û£»
£¨2£©ÁòËáµÄÎïÖʵÄÁ¿Îª$\frac{350g¡Á70%}{98g/mol}$=2.5mol£¬n£¨H+£©=5mol£¬NaOHµÄÎïÖʵÄÁ¿$\frac{240g}{40g/mol}$=6mol£¬
¼×ÖУºÓÉAl¡«OH-¡«AlO2-¡«H+¡«Al£¨OH£©3¡ý£¬6molNaOH·´Ó¦£¬µÃµ½6molAlO2-£¬ÏûºÄ5molH+Éú³É5molAl£¨OH£©3£»
ÒÒÖУºÓÉAl¡«3H+¡«Al3+¡«3OH-¡«Al£¨OH£©3¡ý£¬5molH+·´Ó¦£¬µÃµ½$\frac{5}{3}$molAl3+£¬ÏûºÄ5molOH-µÃµ½$\frac{5}{3}$molAl£¨OH£©3£¬Ê£Óà1molOH-£¬ÓÉAl£¨OH£©3¡ý¡«OH-¡«AlO2-£¬¿ÉÒÔÈܽâ1molAl£¨OH£©3£¬×îÖյõ½Al£¨OH£©3Ϊ$\frac{5}{3}$mol-1mol=$\frac{2}{3}$mol£»
±ûÖУºÓÉAl¡«3H+¡«Al3+£¬¿ÉÖª5molH+·´Ó¦£¬µÃµ½$\frac{5}{3}$molAl3+£¬ÓÉAl¡«OH-¡«AlO2-£¬¿ÉÖª6molNaOH·´Ó¦£¬µÃµ½6molAlO2-£¬ÓÉ3AlO2-¡«Al3+¡«4Al£¨OH£©3¡ý£¬¿ÉÖªAl3+²»×㣬¹ÊµÃµ½Al£¨OH£©3Ϊ$\frac{5}{3}$mol¡Á4=$\frac{20}{3}$mol£¬
¹ÊW1£¬W2£¬W3ÆäÖÊÁ¿ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ£ºW3£¾W1£¾W2£¬
¹Ê´ð°¸Îª£ºW3£¾W1£¾W2£»
£¨3£©¸ù¾Ý£¨2£©ÖмÆËã¿ÉÖª£¬Al£¨OH£©3µÄ×î´óÖÊÁ¿Îª$\frac{20}{3}$mol¡Á78g/mol=520g£¬
¹Ê´ð°¸Îª£º520g£®

µãÆÀ ±¾Ì⿼²éʵÑé·½°¸ÆÀ¼Û¡¢»¯Ñ§·½³ÌʽÓйؼÆË㣬ÌâÄ¿Éæ¼°¹ý³Ì¶à£¬×¢ÒâÀûÓùØÏµÊ½½â´ð£¬×¢ÒâÁòËá¡¢ÇâÑõ»¯ÄÆÎª¶¨Á¿£¬²»Äܸù¾ÝµÃµ½ÏàͬÁ¿µÄÇâÑõ»¯ÂÁÏûºÄÁòËá¡¢ÇâÑõ»¯ÄƵÄÁ¿ÅжϳÁµíÖÊÁ¿¹ØÏµ£¬ÄѶȽϴó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
12£®ÏÂÃæÊÇijͬѧÑо¿ÂÈˮƯ°×ÐÔµÄÒ»¸ö̽¾¿ÐÔʵÑ鯬¶Ï£®
»î¶¯¼ÆÂ¼£º¡¾¹Û²ì¡¿ÂÈÆøµÄÑÕÉ«£º»ÆÂÌÉ«£»ÂÈË®µÄÑÕÉ«£º³Ê»ÆÂÌÉ«£®
¡¾½áÂÛ¡¿ÂÈË®Öк¬ÓÐÂÈÆø·Ö×Ó£®
¡¾ÊµÑé²Ù×÷¡¿ÓÃÁ½¸ù²£Á§°ô·Ö±ðպȡÑÎËáºÍÂÈË®£¬¸÷µÎÔÚÁ½Æ¬À¶É«Ê¯ÈïÊÔÖ½ÉÏ£®
¡¾ÊµÑéÏÖÏó¡¿µÎÓÐÑÎËáµÄÊÔÖ½±äºìÉ«£¬µÎÓÐÂÈË®µÄÊÔÖ½Öмä±ä°×£¬ÍâȦ±äºì£®
¡¾·ÖÎöÓë½áÂÛ¡¿ÂÈË®³Ê»ÆÂÌÉ«£¬ËµÃ÷ÂÈË®ÖÐÈÜÓÐÂÈÆø·Ö×Ó£®µÎÓÐÂÈË®µÄÀ¶É«Ê¯ÈïÊÔÖ½ÍâȦ±äºì£¬ËµÃ÷ÂÈË®ÖÐÓÐÄÜʹÊÔÖ½±äºìµÄËáÉú³É£»Öмä±ä°×£¬ËµÃ÷ÂÈË®ÖÐÓÐÄܹ»Ê¹ÓÐÉ«ÎïÖÊÍÊÉ«µÄÎïÖÊÉú³É£®
¡¾ÎÊÌâÓë˼¿¼¡¿ÂÈÆøÈÜÓÚË®·¢ÉúÈçÏ·´Ó¦£ºCl2+H2O¨THCl+HClO£¬ÈÜÒºÖеÄË®ºÍÑÎËᶼûÓÐÆ¯°××÷Óã¬Äܹ»Ê¹ÓÐÉ«ÎïÖÊÍÊÉ«µÄÎïÖÊÊÇÂÈË®ÖеÄÂÈÆø·Ö×Ó»¹ÊÇÂÈË®ÖеĴÎÂÈËáÄØ£¿»¹ÊǶþÕß¶¼ÓÐÆ¯°××÷Óã¿
ÇëÄã²ÎÕÕËûÒÑÍê³ÉµÄ²¿·ÖʵÑé»î¶¯¼Ç¼£¬×Ô¼ºÉè¼ÆÒ»¸öʵÑ飬֤Ã÷µ½µ×ÊÇÄÄÖÖÎïÖÊÓÐÆ¯°××÷Óã®
¡¾ÊµÑé²Ù×÷¡¿½«¸ÉÔïµÄÂÈÆø·Ö±ðͨÈë×°ÓиÉÔïµÄºìÖ½¡¢³±ÊªµÄºìÖ½µÄ¼¯ÆøÆ¿£®
¡¾ÊµÑéÏÖÏ󡿸ÉÔïµÄºìÖ½²»±äÉ«£¬³±ÊªµÄºìÖ½ÍÊÉ«£®
¡¾·ÖÎöÓë½áÂÛ¡¿ÂÈ·Ö×Ó²»¾ßÓÐÆ¯°××÷Ó㬴ÎÂÈËá²ÅÄÜʹÓÐÉ«ÎïÖÊÍÊÉ«£®

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

¾«Ó¢¼Ò½ÌÍø