ÌâÄ¿ÄÚÈÝ
ʵÑéÊÒÀïÓÃÁòË᳧ÉÕÔü(Ö÷Òª³É·ÖΪÌúµÄÑõ»¯Îï¼°ÉÙÁ¿FeS¡¢SiO2µÈ)ÖƱ¸¾ÛÌú(¼îʽÁòËáÌúµÄ¾ÛºÏÎï)ºÍÂÌ·¯(FeSO4¡¤7H2O)£¬Æä¹ý³ÌÈçÏ£º
(1)¹ý³Ì¢ÙÖУ¬FeSºÍO2¡¢H2SO4·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ ¡£
(2)¹ý³Ì¢ÚÖвúÉúµÄβÆø»á¶Ô´óÆøÔì³ÉÎÛȾ£¬¿ÉÑ¡ÓÃÏÂÁÐÊÔ¼ÁÖÐµÄ ÎüÊÕ¡£
a£®Å¨H2SO4 b£®ÕôÁóË® c£®NaOHÈÜÒº d£®Å¨ÏõËá
(3)¹ý³Ì¢ÛÖУ¬ÐèÒª¼ÓÈëµÄÎïÖÊÃû³ÆÊÇ
(4)¹ý³Ì¢ÜµÄʵÑé²Ù×÷ÊÇ
(5)¹ý³Ì¢ÞÖУ¬½«ÈÜÒºZ¼ÓÈȵ½70¡«80¡æ£¬Ä¿µÄÊÇ ¡£
(6)ʵÑéÊÒΪ²â¶¨ËùµÃµ½µÄ¾ÛÌúÑùÆ·ÖÐÌúÔªËصÄÖÊÁ¿·ÖÊý£¬½øÐÐÏÂÁÐʵÑé¡£¢ÙÓ÷ÖÎöÌìƽ³ÆÈ¡ÑùÆ·2.700 g£»¢Ú½«ÑùÆ·ÈÜÓÚ×ãÁ¿ÑÎËáºó£¬¼ÓÈË×ãÁ¿µÄÂÈ»¯±µÈÜÒº£»¢Û¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔ³ÆÁ¿£¬µÃ¹ÌÌåÖÊÁ¿Îª3.495 g¡£Èô¸Ã¾ÛÌúÖ÷Òª³É·ÖΪ[Fe(OH)SO4]n£¬Ôò¸Ã¾ÛÌúÖÐÌúÔªËصÄÖÊÁ¿·ÖÊýΪ ¡£
(1)¹ý³Ì¢ÙÖУ¬FeSºÍO2¡¢H2SO4·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ ¡£
(2)¹ý³Ì¢ÚÖвúÉúµÄβÆø»á¶Ô´óÆøÔì³ÉÎÛȾ£¬¿ÉÑ¡ÓÃÏÂÁÐÊÔ¼ÁÖÐµÄ ÎüÊÕ¡£
a£®Å¨H2SO4 b£®ÕôÁóË® c£®NaOHÈÜÒº d£®Å¨ÏõËá
(3)¹ý³Ì¢ÛÖУ¬ÐèÒª¼ÓÈëµÄÎïÖÊÃû³ÆÊÇ
(4)¹ý³Ì¢ÜµÄʵÑé²Ù×÷ÊÇ
(5)¹ý³Ì¢ÞÖУ¬½«ÈÜÒºZ¼ÓÈȵ½70¡«80¡æ£¬Ä¿µÄÊÇ ¡£
(6)ʵÑéÊÒΪ²â¶¨ËùµÃµ½µÄ¾ÛÌúÑùÆ·ÖÐÌúÔªËصÄÖÊÁ¿·ÖÊý£¬½øÐÐÏÂÁÐʵÑé¡£¢ÙÓ÷ÖÎöÌìƽ³ÆÈ¡ÑùÆ·2.700 g£»¢Ú½«ÑùÆ·ÈÜÓÚ×ãÁ¿ÑÎËáºó£¬¼ÓÈË×ãÁ¿µÄÂÈ»¯±µÈÜÒº£»¢Û¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔ³ÆÁ¿£¬µÃ¹ÌÌåÖÊÁ¿Îª3.495 g¡£Èô¸Ã¾ÛÌúÖ÷Òª³É·ÖΪ[Fe(OH)SO4]n£¬Ôò¸Ã¾ÛÌúÖÐÌúÔªËصÄÖÊÁ¿·ÖÊýΪ ¡£
£¨1£©4FeS+3O2+6H2SO4=2Fe2(SO4)3+6H2O+4S£¨2·Ö£©
£¨2£©c£¨2·Ö£© £¨3£©Ìú£¨2·Ö£©
£¨4£©Õô·¢£¨Å¨Ëõ£©¡¢£¨ÀäÈ´£©½á¾§¡¢¹ýÂË¡¢Ï´µÓ£¨4·Ö£¬Å¨Ëõ¡¢ÀäÈ´²»Ð´¿ÉÒÔ²»¿Û·Ö£©
£¨5£©´Ù½øFe3+µÄË®½â£¨2·Ö£©
£¨6£©31.11%£¨3·Ö£©
ÊÔÌâ·ÖÎö£º
£¨1£©¸ù¾Ý¹ÌÌåWµÄ³É·ÖÖдæÔÚS£¬ÍƳöFeSºÍO2¡¢H2SO4·¢ÉúÑõ»¯»¹Ô·´Ó¦£¬FeS×ö»¹Ô¼Á£¬O2×÷Ñõ»¯¼Á£¬²úÎïÓÐFe2(SO4)3¡¢S¡¢H2O¡£
£¨2£©ÎÛȾÎïÊÇSO2£¬ÓüîÒºÎüÊÕ¡£
£¨3£©ÈÜÒºXÖеÄÌúÊÇFe3+£¬ÂÌ·¯ÖеÄÌúΪFe2+£¬¹ÊÓ¦¼ÓÈëFe·Û£¬½«Fe3+»¹ÔΪFe2+¡£
£¨4£©¶ÔÄÜÐγɽᾧˮºÏÎïµÄÑÎÈÜÒº²»ÄܲÉÈ¡Ö±½ÓÕô¸ÉµÄ°ì·¨ÖÆÈ¡¾§Ìå¡£
£¨5£©Éý¸ßζȣ¬´Ù½øFe3+µÄË®½â¡£ £¨6£©¢ÛµÃµ½µÄ³ÁµíÊÇBaSO4£¬n(BaSO4) ==0.015mol£¬¸ù¾Ý¾ÛÌúÖ÷Òª³É·ÖΪ[Fe(OH)SO4] n£¬ÍƳön(Fe3+)=0.015mol£¬Ôòm(Fe3+)=0.84g£¬ÌúÔªËصÄÖÊÁ¿·ÖÊý:¦Ø(Fe)=(0.84g/2.700g)¡Á100%=31.11%¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿