ÌâÄ¿ÄÚÈÝ
Ò»°ü¹ÌÌå·ÛÄ©£¬Ëü¿ÉÄÜÊÇÓÉNaCl¡¢CuSO4¡¢Na2O2¡¢K2CO3¡¢£¨NH4£©2SO4¡¢Na2SO4¡¢KNO3µÈÆßÖÖÎïÖÊÖеÄij¼¸ÖÖ»ìºÏ¶ø³É¡£Ä³Í¬Ñ§¶ÔËü½øÐÐÁËÈçϲÙ×÷£¬È»ºó¾¹ý·ÖÎöºÍ¼ÆËãÈ·¶¨ÁËËüµÄ×é³ÉºÍ¸÷³É·ÖµÄÎïÖʵÄÁ¿¡£¢ñ.³ÆÈ¡
¢ò.½«¢ñÖÐËùµÃÈÜÒº·Ö³ÉÁ½µÈ·Ý£¬Ò»·Ý¼ÓÈë·Ó̪ÈÜÒº³ÊºìÉ«£¬ÓÃŨ¶ÈΪ4.0 mol¡¤L-1?µÄÑÎËáÖкͣ¬ÓÃÈ¥¸ÃÑÎËá12.5 mL¡£½«ÁíÒ»·ÝÓÃÏ¡ÏõËáËữ£¬ÎÞÆøÌå²úÉú£¬ÔÙ¼ÓÈë¹ýÁ¿µÄBa£¨NO3£©2ÈÜÒº£¬µÃ°×É«³Áµí£¬ÖÊÁ¿Îª
¢ó.½«¢ñÖÐËùµÃµÄ
£¨1£©¶¨ÐÔ·ÖÎö£ºÓÉÉÏÊöʵÑéÏÖÏ󣨲»½øÐÐÊý¾ÝµÄ¼ÆË㣩¿ÉÒÔÈ·¶¨¸Ã·ÛÄ©ÖÐÒ»¶¨²»º¬ÓеÄÊÇ__________£»Ò»¶¨º¬ÓеÄÊÇ__________¡£
£¨2£©¶¨Á¿¼ÆË㣺ΪÁ˽øÒ»²½È·¶¨¸Ã·ÛÄ©ÖÐÆäËû³É·ÖÊÇ·ñ´æÔÚ£¬¸ÃͬѧÓÖ¶ÔÓйØÊý¾Ý½øÐÐÁË´¦Àí£¬²¢×îÖյóö»ìºÏÎïµÄ×é³ÉºÍ¸÷³É·ÖµÄÎïÖʵÄÁ¿£¬Çëд³öËûµÄ¼ÆËã¹ý³Ì¡£
£¨1£©CuSO4¡¢K2CO3¡¢NaCl£»Na2O2¡¢£¨NH4£©2SO4
£¨2£©Óɶ¨ÐÔ·ÖÎöÖª£¬Na2O2¡¢£¨NH4£©2SO4Ò»¶¨º¬ÓУ¬Na2SO4¡¢KNO3¿ÉÄܺ¬ÓÐ
Éè
ÓÉ2Na2O2+2H2O====4NaOH+O2¡ü
x 2x x/2
2NaOH+£¨NH4£©2SO4====Na2SO4+2NH3¡ü+2H2O
2y y 2y
ËùÒÔ£º+2y==0.15 mol¢Ù
ÓÉÌâÒâÖª£¬ÉÏÊö·´Ó¦ÖÐNaOH¹ýÁ¿£¬¹ýÁ¿µÄNaOHÓÃÑÎËáÖкͣ»
ËùÒÔ£º2x-2y=2¡Á0.012
×ۺϣ¬½âµÃx=0.1 mol£»y=0.05 mol
½«ÈÜÒº·Ö³ÉÁ½µÈ·Ýºó£¬ÏòÒ»·ÝÖмÓÈë¹ýÁ¿µÄBa£¨NO3£©2ÈÜÒº£¬¿ÉÒÔÉú³ÉBaSO4µÄÎïÖʵÄÁ¿Îª£º=0.055 mol
0.1 mol¡Á
ËùÒÔ
½âÎö£º·ÛÄ©ÈÜÓÚÕôÁóË®µÃÎÞÉ«ÈÜÒº£¬ËµÃ÷ÎÞCuSO4¡£²úÉúÆøÌ壬ÆøÌåͨ¹ýŨÁòËᣬÆøÌåÌå»ýÃ÷ÏÔ¼õС£¬ËµÃ÷ÆøÌåÖÐÓÐNH3£¬ÔÙͨ¹ý¼îʯ»Ò£¬ÆøÌåÌå»ý²»±ä£¬ËµÃ÷ÓÐO2£¬´Ó¶ø˵Ã÷ÔÑùÆ·Ò»¶¨º¬£¨NH4£©2SO4ºÍNa2O2¡£ÓÃÏõËáËữÎÞÆøÌå²úÉú£¬ËµÃ÷ÎÞNa2CO3£¬¼ÓBa£¨NO3£©2ºóÔÙ¼ÓÈëAgNO3ÎÞÃ÷ÏÔÏÖÏó£¬ËµÃ÷ÎÞNaCl¡£Na2SO4¡¢KNO3ÎÞ·¨Åжϡ£
£¨2£©¼û´ð°¸¡£