Question

17£®ÁòÓжàÖÖ»¯ºÏÎï
£¨1£©Áò»¯Çâ¾ßÓл¹Ô­ÐÔ£¬¿ÉÒÔºÍÐí¶àÑõ»¯¼Á·´Ó¦£®H₂SͨÈëÁòËáËữµÄKMnO₄ÈÜÒº£¬·´Ó¦Éú³ÉS¡¢MnSO₄µÈ£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º3H₂SO₄+5H₂S+2KMnO₄=5S¡ý+2MnSO₄+K₂SO₄+8H₂O£®
£¨2£©½«ºÚÉ«µÄFe₂S₃¹ÌÌå¼ÓÈë×ãÁ¿µÄÑÎËáÖУ¬ÈÜÒºÖÐÓе­»ÆÉ«¹ÌÌåÉú³É£¬²úÎﻹÓÐÂÈ»¯ÑÇÌú£¬Áò»¯Ç⣮
£¨3£©ÁòÌú¿óÊǹ¤ÒµÉú²úÁòËáµÄÖ÷ÒªÔ­ÁÏ£¬ÁòÌú¿óÑõ»¯±ºÉյĻ¯Ñ§·´Ó¦ÈçÏ£º
3FeS₂+8O₂¡úFe₃O₄+6SO₂     4FeS₂+11O₂¡ú2Fe₂O₃+8SO₂
Èô48mol FeS₂ÍêÈ«·´Ó¦ºÄÓÃÑõÆø2934.4L£¨±ê×¼×´¿ö£©£»·´Ó¦²úÎïÖÐFe₃O₄ÓëFe₂O₃ÎïÖʵÄÁ¿Ö®±ÈΪ2£º9£®

Answer 1

·ÖÎö £¨1£©H₂SͨÈëÁòËáËữµÄKMnO₄ÈÜÒº£¬·´Ó¦Éú³ÉS¡¢MnSO₄µÈ£¬SÔªËصĻ¯ºÏ¼ÛÉý¸ß£¬MnÔªËصĻ¯ºÏ¼Û½µµÍ£¬½áºÏµç×Ó¡¢Ô­×ÓÊغã·ÖÎö£»
£¨2£©ºÚÉ«µÄFe₂S₃¹ÌÌå¼ÓÈë×ãÁ¿µÄÑÎËáÖУ¬ÈÜÒºÖÐÓе­»ÆÉ«¹ÌÌåÉú³É£¬µ­»ÆÉ«¹ÌÌåΪS£¬¿ÉÖªSÔªËصĻ¯ºÏ¼ÛÉý¸ß£¬ÔòFeÔªËصĻ¯ºÏ¼Û½µµÍ£»
£¨3£©n£¨O₂£©=$\frac{2934.4L}{22.4L/mol}$=131mol£¬Éè·´Ó¦²úÎïÖÐFe₃O₄ÓëFe₂O₃ÎïÖʵÄÁ¿·Ö±ðΪx¡¢y£¬½áºÏ·´Ó¦¼ÆË㣮

½â´ð ½â£º£¨1£©H₂SͨÈëÁòËáËữµÄKMnO₄ÈÜÒº£¬·´Ó¦Éú³ÉS¡¢MnSO₄µÈ£¬SÔªËصĻ¯ºÏ¼ÛÉý¸ß£¬MnÔªËصĻ¯ºÏ¼Û½µµÍ£¬Óɵç×Ó¡¢Ô­×ÓÊغã¿ÉÖª·´Ó¦Îª3H₂SO₄+5H₂S+2KMnO₄=5S¡ý+2MnSO₄+K₂SO₄+8H₂O£¬¹Ê´ð°¸Îª£º3H₂SO₄+5H₂S+2KMnO₄=5S¡ý+2MnSO₄+K₂SO₄+8H₂O£»
£¨2£©ºÚÉ«µÄFe₂S₃¹ÌÌå¼ÓÈë×ãÁ¿µÄÑÎËáÖУ¬ÈÜÒºÖÐÓе­»ÆÉ«¹ÌÌåÉú³É£¬µ­»ÆÉ«¹ÌÌåΪS£¬¿ÉÖªSÔªËصĻ¯ºÏ¼ÛÉý¸ß£¬ÔòFeÔªËصĻ¯ºÏ¼Û½µµÍ£¬½áºÏH¡¢ClÔªËØÊغã¿ÉÖª£¬·´Ó¦²úÎﻹÓÐÂÈ»¯ÑÇÌú¡¢Áò»¯Ç⣬¹Ê´ð°¸Îª£ºÂÈ»¯ÑÇÌú£»Áò»¯Ç⣻
£¨3£©n£¨O₂£©=$\frac{2934.4L}{22.4L/mol}$=131mol£¬Éè·´Ó¦²úÎïÖÐFe₃O₄ÓëFe₂O₃ÎïÖʵÄÁ¿·Ö±ðΪx¡¢y£¬Ôò
ÓÉ3FeS₂+8O₂¡úFe₃O₄+6SO₂¡¢4FeS₂+11O₂¡ú2Fe₂O₃+8SO₂¿ÉÖª£¬
$\left\{\begin{array}{l}{3x+2y=48}\\{8x+5.5y=131}\end{array}\right.$£¬
½âµÃx=4mol£¬y=18mol£¬
ËùÒÔFe₃O₄ÓëFe₂O₃ÎïÖʵÄÁ¿Ö®±È4£º18=2£º9£¬
¹Ê´ð°¸Îª£º2£º9£®

µãÆÀ ±¾Ì⿼²éÑõ»¯»¹Ô­·´Ó¦¼°¼ÆË㣬Ϊ¸ß¿¼¸ßƵ¿¼µã£¬°ÑÎÕ·¢ÉúµÄ»¯Ñ§·´Ó¦¼°·´Ó¦ÖÐÔªËصĻ¯ºÏ¼Û±ä»¯Îª½â´ðµÄ¹Ø¼ü£¬²àÖØ»ù±¾¸ÅÄîºÍ¼ÆËãÄÜÁ¦µÄ¿¼²é£¬ÌâÄ¿ÄѶȲ»´ó£®