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Answer 1

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1£©FeS₂

£¨2£©

£¨3£©a   b   d

£¨4£©SO₃(g) + H₂O(l) = H₂SO₄(l)£»¦¤H=-130.3kJ/mol

£¨5£©¢ÙSO₂ + Br₂ + 2H₂O = 4H⁺ + 2Br^- + SO₄^2-

¢Ú14.56      2.31g

 

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£¨1£©¸ù¾ÝÔ­×ÓÊغ㼴¿ÉÅжϸÃÎïÖÊΪFeS₂¡£

£¨2£©SO₂ºÍO₂ÔÚ½Ó´¥ÊÒ·¢Éú·´Ó¦Éú³ÉSO₃£º2SO₂ + O₂  2SO₃¡£

£¨3£©Ñ¡Ïîa£¬½«»ÆÌú¿ó·ÛËé¸üÔö¼Ó·´Ó¦½Ó´¥Ãæ»ý¶ø¼Ó¿ì·´Ó¦ËÙÂÊ¡£Ñ¡Ïîb£¬Ôö¼Ó¿ÕÆø£¨O₂£©Å¨¶ÈÄÜÌá¸ßSO₂µÄת»¯ÂÊ¡£Ñ¡Ïîc£¬´ß»¯¼Á¶Ôת»¯ÂÊÎÞÓ°Ï졣ѡÏîd£¬¿óÔüÖк¬ÓÐFe₂O₃£¬¿ÉÓÃÓÚÒ±Á¶Ìú¡£

£¨4£©160g SO₃(g)ÓëH₂O(l)·´Ó¦Éú³ÉH₂SO₄(l)·Å³ö260.6kJÈÈÁ¿£¬Ôò1mol SO₃(g)ÓëH₂O(l)·´Ó¦Éú³ÉH₂SO₄(l)·Å³ö130.3kJÈÈÁ¿£¬ÔòÏà¹ØÈÈ»¯Ñ§·½³ÌʽΪ£º SO₃(g) + H₂O(l)£½H₂SO₄(l)£»¡÷H£½£­130.3kJ/mol¡£

£¨5£©10gʱ˵Ã÷ï§Ñβ»×ã,20gʱ˵Ã÷ÇâÑõ»¯ÄƲ»×ã,ÓÉ10.00gºÍ20.00g ʱŨÁòËáÔö¼ÓµÄÖÊÁ¿Ïàͬ˵Ã÷ï§ÑÎÓÐËáʽ¸ù²úÉúµÄ°±Æø, ÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿Å¨¶ÈΪCmol/L

Éè10¿ËÖÐÓÐXmolµÄ (NH₄)₂SO₄ ºÍymolµÄNH₄HSO₄Ôò

132X + 115y=10¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­.(1)

10gʱ˵Ã÷ï§Ñβ»×ãÓÉNÊغãÖª

          n(NH₃)=2X + y

20gʱï§ÑÎÒѾ­¹ýÁ¿£¬ÇâÑõ»¯ÄÆÏȺÍËáʽ¸ù·´Ó¦£¬Ôò(NH₄)₂SO₄Ϊ2Xmol£¬2ymolµÄNH₄HSO₄

        HSO₄^¡ª + OH^¡ª  = SO₄^2¡ª + H₂O

1       1

2y     2y

        NH₄⁺  +   OH^¡ª     =     NH₃       +          H₂O

                  1            1

              50C¡Á10^¡ª3¡ª2y   50C¡Á10^¡ª3¡ª2y

ÓÉÓÚ²úÉúµÄ°±ÆøÒ»Ñù¶à 

        n(NH₃)=2X + y=50C¡Á10^¡ª3¡ª2y¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­.(2)

30gʱï§ÑιýÁ¿£¬Ôò(NH₄)₂SO₄Ϊ3Xmol£¬3ymolµÄNH₄HSO₄   n(NH₃)=0.68/17=0.04mol

HSO₄^¡ª + OH^¡ª  = SO₄^2¡ª + H₂O

1       1

3y      3y

NH₄⁺  +   OH^¡ª        =         NH₃       +   H₂O

                  1                  1

              50C¡Á10^¡ª3¡ª3y         0.04

ËùÒÔ      

50C¡Á10^¡ª3¡ª3y=0.04¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­.(3)

ÁªÁ¢(1) (2) (3)½âµÃ

       X=0.02mol   y=0.064mol    C=4.64mol/L

µªµÄÖÊÁ¿°Ù·Öº¬Á¿= ¡Á100£¥=(0.04+0.064) ¡Á14/10¡Á100£¥=14.56£¥

15gʱï§ÑÎÒѾ­¹ýÁ¿£¬Ôò(NH₄)₂SO₄Ϊ1.5Xmol£¬1.5 ymolµÄNH₄HSO₄

HSO₄^¡ª   +   OH^¡ª  = SO₄^2¡ª + H₂O

1          1

1.5y      1.5y

NH₄⁺  +   OH^¡ª        =         NH₃       +   H₂O

                  1                  1

              50C¡Á10^¡ª3¡ª1.5y     50C¡Á10^¡ª3¡ª1.5y

n(NH₃)= 50C¡Á10^¡ª3¡ª1.5y=50¡Á4.64¡Á10^¡ª3¡ª1.5¡Á0.064=0.136mol

m(NH₃)=0.136¡Á17=2.31g