Question

ÔÚÏÂͼת»¯¹ØϵÖУ¬ÒÑÖªB¡¢D¶¼Êǵ­»ÆÉ«¹ÌÌ壬Çë»Ø´ðÏÂÁÐÎÊÌâ

£¨1£©Ð´³öÏÂÁÐÎïÖʵĻ¯Ñ§Ê½£ºB

 

£¨2£©Ð´³öB¡úC»¯Ñ§·½³Ìʽ

 

£¨3£©½«¹ýÁ¿ÆøÌåEͨÈëÏÂÁи÷×éÈÜÒººó£¬Àë×ÓÈÔÄÜ´óÁ¿¹²´æµÄÊÇ

 

A£®Ca²⁺¡¢ClO^-¡¢Cl^-B£®OH^-¡¢CO₃^2-¡¢Na⁺C£®Ba⁺¡¢Ca²⁺¡¢Cl^-D£®H⁺¡¢Na⁺¡¢NO₃⁺
£¨4£©Ä³Í¬Ñ§Ñо¿ÂÁ¼°Æ仯ºÏÎïµÄÐÔÖÊÊÐÉè¼ÆÁËÈçÏÂÁ½¸öʵÑé·½°¸£º
·½°¸¢Ù£º2.7gAl

100molGÈÜÒº

XÈÜÒº

CÈÜÒº

Al£¨OH£©₂³Áµí
·½°¸¢Ú£º2.7gAl

100molCÈÜÒº

YÈÜÒº

GÈÜÒº

Al£¨OH£©₂³Áµí
ÒÑÖªCÈÜÒººÍGµÄŨ¶È·Ö±ðΪ3mol/LºÍ1.5mol/L£¬ÏòXÈÜÒººÍYÈÜÒºÖзֱð¼ÓÈëCÈÜÒººÍGÈÜҺʱ£¬²úÉú³ÁµíµÄÎïÖʵÄÁ¿Óë¼ÓÈëCÈÜÒººÍGÈÜÒºÌå»ýµÄ¹ØϵÈçͼËùʾ£º

¢Ùд³öÂÁÓëCÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ

 

¢ÚOb¶Î·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ

 

¢ÛOµãʱÁ½·½°¸ÖÐËùµÃÈÜÒºÈÜÖʾùΪNa₂SO₄£¬ÆäÎïÖʵÄÁ¿Å¨¶ÈΪ

 

£®

Answer 1

¿¼µã£ºÎÞ»úÎïµÄÍƶÏ,Àë×Ó·½³ÌʽµÄÓйؼÆËã

רÌ⣺ÍƶÏÌâ,¼ÆËãÌâ

·ÖÎö£º£¨1£©£¨2£©£¨3£©³£¼ûµ­»ÆÉ«¹ÌÌåÓÐNa₂O₂ºÍS£¬ÆäÖÐBÄÜÓëË®·´Ó¦Éú³ÉÑõÆø£¬ÔòBΪNa₂O₂£¬AΪNa£¬CΪNaOH£¬D¿ÉÒÔÓëÑõÆøÁ¬Ðø·´Ó¦£¬ÔòDΪS£¬EΪSO₂£¬FΪSO₃£¬GΪH₂SO₄£¬GÓëNaOH·¢ÉúÖкͷ´Ó¦Éú³ÉHΪNa₂SO₄£»
£¨4£©¢ÙAlÓëÇâÑõ»¯ÄÆ·´Ó¦Éú³ÉÆ«ÂÁËáÄÆ¡¢ÇâÆøÓëË®£»
¢Ún£¨H₂SO₄£©=1.5mol/L¡Á0.1L=0.15mol£¬n£¨NaOH£©=3mol/L¡Á0.1L=0.3mol£¬n£¨Al£©=

2.7g

27g/mol

=0.1mol£¬¸ù¾Ý2Al+3H₂SO₄=Al₂£¨SO₄£©₂+3H₂¡ü£¬AlºÍH₂SO₄ÍêÈ«·´Ó¦ÐèÒªn£¨H₂SO₄£©=

3

2

n£¨Al£©=

3

2

¡Á0.1mol=0.15mol£¬ËùÒÔAlºÍH₂SO₄Ç¡ºÃÍêÈ«·´Ó¦Éú³ÉAl₂£¨SO₄£©₂£¬ÔÙ¼ÓÈëNaOHÈÜÒº»áÁ¢¼´µÃµ½ÇâÑõ»¯ÂÁ³Áµí£»
¸ù¾Ý2Al+2NaOH+2H₂O=2NaAlO₂+3H₂¡ü£¬AlºÍNaOHÍêÈ«·´Ó¦ÐèÒªn£¨NaOH£©=n£¨Al£©=0.1mol£¼0.3mol£¬ËùÒÔNaOHÓÐÊ£Ó࣬ÔÙ¼ÓÈëÁòËᣬÏÈ·¢ÉúÖкͷ´Ó¦H⁺+OH^-=H₂O£¬ÔÙ·¢Éú£ºH⁺+AlO₂^-+H₂O=Al£¨OH£©₃¡ý£»
¢Û¸ù¾ÝÁòËá¸ùÊغã¼ÆËãn£¨H₂SO₄£©£¬ÈÜÒºÌå»ýΪÁòËáÓëÇâÑõ»¯ÄÆÈÜÒºÌå»ýÖ®ºÍ£¬½ø¶ø¼ÆËãÁòËáÄÆÎïÖʵÄÁ¿Å¨¶È£®

½â´ð£º ½â£º³£¼ûµ­»ÆÉ«¹ÌÌåÓÐNa₂O₂ºÍS£¬ÆäÖÐBÄÜÓëË®·´Ó¦Éú³ÉÑõÆø£¬ÔòBΪNa₂O₂£¬AΪNa£¬CΪNaOH£¬D¿ÉÒÔÓëÑõÆøÁ¬Ðø·´Ó¦£¬ÔòDΪS£¬EΪSO₂£¬FΪSO₃£¬GΪH₂SO₄£¬GÓëNaOH·¢ÉúÖкͷ´Ó¦Éú³ÉHΪNa₂SO₄£»
£¨1£©ÓÉÒÔÉÏ·ÖÎö¿ÉÖª£¬BΪNa₂O₂£¬¹Ê´ð°¸Îª£ºNa₂O₂£»
£¨2£©B¡úCµÄ·´Ó¦Îª2Na₂O₂+2H₂O=4NaOH+O₂¡ü£¬¹Ê´ð°¸Îª£º2Na₂O₂+2H₂O=4NaOH+O₂¡ü£»
£¨3£©EΪSO₂£¬¿ÉÓëOH^-¡¢ClO^-¡¢NO₃^-£¨ËáÐÔ£©£¬ÔòÀë×ÓÈÔÄÜ´óÁ¿¹²´æµÄÊÇC£¬¹Ê´ð°¸Îª£ºC£»
£¨4£©¢ÙÂÁÓëNaOHÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º2Al+2OH^-+2H₂O=2AlO₂^-+3H₂¡ü£¬¹Ê´ð°¸Îª£º2Al+2OH^-+2H₂O=2AlO₂^-+3H₂¡ü£»
¢Ún£¨H₂SO₄£©=1.5mol/L¡Á0.1L=0.15mol£¬n£¨NaOH£©=3mol/L¡Á0.1L=0.3mol£¬n£¨Al£©=

2.7g

27g/mol

=0.1mol£¬¸ù¾Ý2Al+3H₂SO₄=Al₂£¨SO₄£©₂+3H₂¡ü£¬AlºÍH₂SO₄ÍêÈ«·´Ó¦ÐèÒªn£¨H₂SO₄£©=

3

2

n£¨Al£©=

3

2

¡Á0.1mol=0.15mol£¬ËùÒÔAlºÍH₂SO₄Ç¡ºÃÍêÈ«·´Ó¦Éú³ÉAl₂£¨SO₄£©₂£¬ÔÙ¼ÓÈëNaOHÈÜÒº»áÁ¢¼´µÃµ½ÇâÑõ»¯ÂÁ³Áµí£»
¸ù¾Ý2Al+2NaOH+2H₂O=2NaAlO₂+3H₂¡ü£¬AlºÍNaOHÍêÈ«·´Ó¦ÐèÒªn£¨NaOH£©=n£¨Al£©=0.1mol£¼0.3mol£¬ËùÒÔNaOHÓÐÊ£Ó࣬ÔÙ¼ÓÈëÁòËᣬÏÈ·¢ÉúÖкͷ´Ó¦H⁺+OH^-=H₂O£¬ÔÙ·¢Éú£ºH⁺+AlO₂^-+H₂O=Al£¨OH£©₃¡ý£¬
¹ÊOb¶Î·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºH⁺+AlO₂^-+H₂O=Al£¨OH£©₃¡ý£¬
¹Ê´ð°¸Îª£ºH⁺+AlO₂^-+H₂O=Al£¨OH£©₃¡ý£»
¢ÛOµãʱÁ½·½°¸ÖÐËùµÃÈÜÒºÈÜÖʾùΪNa₂SO₄£¬ÈÜÒºÌå»ýΪÁòËáÓëÇâÑõ»¯ÄÆÈÜÒºÌå»ýÖ®ºÍ£¬¸ù¾ÝÁòËá¸ùÊغãn£¨Na₂SO₄£©=n£¨H₂SO₄£©=0.15mol£¬Ôòc£¨Na₂SO₄£©=

0.15mol

0.2L

=0.75mol/mol£¬
¹Ê´ð°¸Îª£º0.75mol/L£®

µãÆÀ£º±¾Ì⿼²éÎÞ»úÎïµÄÍƶϡ¢ÐÔÖÊʵÑé·½°¸Éè¼Æ£¬£¨4£©ÖйؼüÊǸù¾Ý¹ýÁ¿¼ÆËãÃ÷È·¸÷Ï߶η¢ÉúµÄ»¯Ñ§·´Ó¦£¬²àÖØ¿¼²éѧÉú·ÖÎöÎÊÌâ¡¢½â´ðÎÊÌâÄÜÁ¦£¬×¢ÒâÏà¹Ø»ù´¡ÖªÊ¶µÄ»ýÀÛ£®