Question

½«0.2mol Mg¡¢AlµÄ»ìºÏÎïÓÃÑÎËáÍêÈ«Èܽ⣬·Å³ö±ê×¼×´¿öÏÂ5.6LÆøÌ壬ȻºóÔÙµÎÈë2mol?L^-1 NaOHÈÜÒºV mL£¬Çë»Ø´ð£º
£¨1£©Çó¸Ã»ìºÏÎïÖÐMgÖÊÁ¿°Ù·ÖÊýΪ£º

 

%£¨Ð¡Êýµãºó±£Áôһ룩£®
£¨2£©Èô¸Ã»ìºÏÎïÓëÑÎËáÇ¡ºÃÍêÈ«·´Ó¦£¬ÔڵμÓNaOHÈÜÒº¹ý³ÌÖУ¬ÓûʹMg²⁺¡¢Al³⁺¸ÕºÃ³ÁµíÍêÈ«£¬ÔòµÎÈëNaOHÈÜÒºµÄÌå»ýV=

 

mL£®
£¨3£©Èô¸Ã»ìºÏÎïÈÜÓÚ200mL 3mol/LÑÎËáºó£¬µÎÈëNaOHÈÜÒº£¬Ê¹ËùµÃ³ÁµíÖÐÎÞAl£¨OH£©₃£¬ÔòµÎÈëNaOHÈÜÒºµÄÌå»ýV×îÉÙΪ£º

 

mL£®

Answer 1

¿¼µã£ºÓйػìºÏÎï·´Ó¦µÄ¼ÆËã,þ¡¢ÂÁµÄÖØÒª»¯ºÏÎï

רÌ⣺¼ÆËãÌâ

·ÖÎö£º£¨1£©ÉèºÏ½ðÖÐMgΪxmol£¬AlΪymol£¬¸ù¾Ý¶þÕßÖÊÁ¿¼°Éú³ÉÇâÆøµÄÌå»ýÁз½³Ì¼ÆË㣬½ø¶ø¼ÆËãMgµÄÖÊÁ¿·ÖÊý£»
£¨2£©·¢Éú·´Ó¦£ºMg²⁺+20H^-=Mg£¨OH£©₂¡ý¡¢Al³⁺+30H^-=Al£¨OH£©₃¡ý£¬¸ù¾Ý·½³Ìʽ¼ÆËãÐèÒªÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿£¬ÔÙ¸ù¾ÝV=

n

c

¼ÆË㣻
£¨3£©Èô¸ÃMg¡¢Al»ìºÏÎïÈÜÓÚ200mL 3mol/LÑÎËáºó£¬µÎÈëNaOHÈÜÒº£¬Ê¹ËùµÃ³ÁµíÖÐÎÞAl£¨OH£©₃£¬ËµÃ÷AlÔªËØÒÔÆ«ÂÁËáÄÆÐÎʽ´æÔÚ£¬¹Ê´ËʱÈÜÒºÖÐÈÜÖÊΪNaCl¡¢NaAlO₂£¬¸ù¾ÝÄÆÀë×ÓÊغã¿ÉÖªn£¨NaOH£©=n£¨NaCl£©+n£¨NaAlO₂£©£¬ÓÉÂÈÀë×ÓÊغãn£¨NaCl£©=n£¨HCl£©£¬ÓÉAlÔªËØÊغãn£¨NaAlO₂£©=n£¨Al£©£¬¾Ý´Ë¼ÆËãn£¨NaOH£©£¬ÔÙ¸ù¾ÝV=

n

c

¼ÆË㣮

½â´ð£º ½â£º£¨1£©Ã¾ÂÁºÏ½ðÓëÑÎËá·¢Éú£ºMg+2HCl=MgCl₂+H₂¡ü¡¢2Al+6HCl=2AlCl₃+3H₂¡ü£¬ÉèºÏ½ðÖÐMgΪxmol£¬AlΪymol£¬Ôò

x+y=0.2 x+1.5y= 5.6 22.4

£¬½âÖ®µÃx=0.1£¬y=0.1£¬
¹Ê»ìºÏÎïÖÐMgµÄÖÊÁ¿=0.1mol¡Á24g/mol=2.4g£¬AlµÄÖÊÁ¿=0.1mol¡Á27g/mol=2.7g£¬
¹Ê»ìºÏÎïÖÐMgµÄÖÊÁ¿·ÖÊý=

2.4g

2.4g+2.7g

¡Á100%=47.1%£¬
¹Ê´ð°¸Îª£º47.1%£»
£¨2£©Èô¸ÃMg¡¢AlµÄ»ìºÏÎïÓëÑÎËáÇ¡ºÃÍêÈ«·´Ó¦£¬ÔڵμÓNaOHÈÜÒº¹ý³ÌÖУ¬ÓûʹMg²⁺¡¢Al³⁺¸ÕºÃ³ÁµíÍêÈ«£¬
Ôò·¢Éú£ºMg²⁺+20H^-=Mg£¨OH£©₂¡ý¡¢Al³⁺+30H^-=Al£¨OH£©₃¡ý£¬
ÔòÐèÒªn£¨NaOH£©=2n£¨Mg£©+3n£¨Al£©=2¡Á0.1mol+3¡Á0.1mol=0.5mol£¬
V£¨NaOH£©=

0.5mol

2mol/L

=0.25L=250mL£¬
¹Ê´ð°¸Îª£º250£»
£¨3£©Èô¸ÃMg¡¢Al»ìºÏÎïÈÜÓÚ200mL 3mol/LÑÎËáºó£¬µÎÈëNaOHÈÜÒº£¬Ê¹ËùµÃ³ÁµíÖÐÎÞAl£¨OH£©₃£¬ËµÃ÷AlÔªËØÒÔÆ«ÂÁËáÄÆÐÎʽ´æÔÚ£¬¹Ê´ËʱÈÜÒºÖÐÈÜÖÊΪNaCl¡¢NaAlO₂£¬
¸ù¾ÝÄÆÀë×ÓÊغã¿ÉÖªn£¨NaOH£©=n£¨NaCl£©+n£¨NaAlO₂£©£¬ÓÉÂÈÀë×ÓÊغãn£¨NaCl£©=n£¨HCl£©£¬ÓÉAlÔªËØÊغãn£¨NaAlO₂£©=n£¨Al£©£¬¹Ên£¨NaOH£©=n£¨HCl£©+n£¨Al£©=0.2L¡Á3mol/L+0.1mol=0.7mol£¬¹ÊV£¨NaOH£©=

0.7mol

2mol/L

=0.35L=350mL£¬
¹Ê´ð°¸Îª£º350£®

µãÆÀ£º±¾Ì⿼²é»¯Ñ§·½³ÌʽµÄ¼ÆËã¡¢»ìºÏÎï¼ÆË㣬²àÖØÓÚѧÉúµÄ·ÖÎöÄÜÁ¦ºÍ¼ÆËãÄÜÁ¦µÄ¿¼²é£¬£¨3£©×¢Òâ¸ù¾ÝÔªËØÊغã¼ÆË㣬ÄѶÈÖеȣ®