Question

17£®Ïò20.0gCu ºÍCu₂SµÄ»ìºÏÎïÖмÓÈëijŨ¶ÈµÄÏ¡ÏõËá0.5L£¬¹ÌÌåÎïÖÊÍêÈ«·´Ó¦£¬Éú³ÉSµ¥ÖÊ¡¢Cu£¨NO₃£©₂ºÍ±ê×¼×´¿öÏÂ4.48LNO£®¹ýÂ˺óÏòËùµÃÈÜÒºÖмÓÈë1.0mol/LµÄNaOHÈÜÒº1.0L£¬½ðÊôÀë×ÓÍêÈ«³Áµí£¬´ËʱÈÜÒº³ÊÖÐÐÔ£®ÏÂÁÐÓйØ˵·¨²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

• A£® CuÓëCu₂SµÄÎïÖʵÄÁ¿Ö®±ÈΪ10£º1
• B£® ÏõËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ1.2mol/L
• C£® Cu£¨OH£©₂³ÁµíÖÊÁ¿Îª29.4g
• D£® Cu¡¢CuSÓëÏõËá·´Ó¦ºóÊ£ÓàHNO₃Ϊ0.4mol

Answer 1

·ÖÎö ÔÚËùµÃÈÜÒºÖмÓÈë1.0mol/L µÄNaOHÈÜÒº1.0L£¬´ËʱÈÜÒº³ÊÖÐÐÔ£¬½ðÊôÀë×ÓÒÑÍêÈ«³Áµí£¬ÈÜÒºÖÐÈÜÖÊΪNaNO₃£¬n£¨NaNO₃£©=n£¨NaOH£©=1.0mol/L¡Á1.0L=1mol£¬³ÁµíΪCu£¨OH£©₂£¬¹ÌÌåÎïÖÊÍêÈ«·´Ó¦£¬Éú³ÉSµ¥ÖÊ¡¢Cu£¨NO₃£©₂ºÍ±ê×¼×´¿öÏÂ4.48LNO£¬n£¨NO£©=$\frac{4.48L}{22.4L/mol}$=0.2mol£¬½áºÏÑõ»¯»¹Ô­·´Ó¦µ¥ÖÊÊغãÁÐʽ¼ÆË㣬Éè20.0gCu ºÍCu₂SµÄ»ìºÏÎïÖÐÍ­ÎïÖʵÄÁ¿Îªx£¬Ñõ»¯ÑÇÍ­ÎïÖʵÄÁ¿Îªy£¬
Cu¡«Cu²⁺¡«2e^-
 x               2x
Cu₂S¡«2Cu²⁺¡«2e^-
  y                  2y
Cu₂S¡«S¡«2e^-
 y             2y
HNO₃¡«NO¡«3e^-
             1         3
       0.2mol      0.6mol
Ôò64x+160y=20.0
2x+2y+2y=0.6mol
x=0.25mol
y=0.025mol
¸ù¾ÝÍ­ÔªËØÊغãÓÐn£¨Cu£©+2n£¨Cu₂O£©=n[Cu£¨OH£©₂]£¬ËùÒÔ·´Ó¦ºóµÄÈÜÒºÖÐn[[Cu£¨NO₃£©₂]=n[Cu£¨OH£©₂]=0.3mol£¬¹Ê·´Ó¦ÖÐÏõËáÓÐÊ£Ó࣬¾Ý´Ë·ÖÎö¼ÆËãÑ¡ÏîÅжϣ®

½â´ð ½â£ºÔÚËùµÃÈÜÒºÖмÓÈë1.0mol/L µÄNaOHÈÜÒº1.0L£¬´ËʱÈÜÒº³ÊÖÐÐÔ£¬½ðÊôÀë×ÓÒÑÍêÈ«³Áµí£¬ÈÜÒºÖÐÈÜÖÊΪNaNO₃£¬n£¨NaNO₃£©=n£¨NaOH£©=1.0mol/L¡Á1.0L=1mol£¬³ÁµíΪCu£¨OH£©₂£¬¹ÌÌåÎïÖÊÍêÈ«·´Ó¦£¬Éú³ÉSµ¥ÖÊ¡¢Cu£¨NO₃£©₂ºÍ±ê×¼×´¿öÏÂ4.48LNO£¬n£¨NO£©=$\frac{4.48L}{22.4L/mol}$=0.2mol£¬½áºÏÑõ»¯»¹Ô­·´Ó¦µ¥ÖÊÊغãÁÐʽ¼ÆË㣬Éè20.0gCu ºÍCu₂SµÄ»ìºÏÎïÖÐÍ­ÎïÖʵÄÁ¿Îªx£¬Ñõ»¯ÑÇÍ­ÎïÖʵÄÁ¿Îªy£¬
Cu¡«Cu²⁺¡«2e^-
 x               2x
Cu₂S¡«2Cu²⁺¡«2e^-
  y                  2y
Cu₂S¡«S¡«2e^-
 y             2y
HNO₃¡«NO¡«3e^-
             1         3
       0.2mol      0.6mol
Ôò64x+160y=20.0
2x+2y+2y=0.6mol
x=0.25mol
y=0.025mol
¸ù¾ÝÍ­ÔªËØÊغãÓÐn£¨Cu£©+2n£¨Cu₂O£©=n[Cu£¨OH£©₂]£¬ËùÒÔ·´Ó¦ºóµÄÈÜÒºÖÐn[[Cu£¨NO₃£©₂]=n[Cu£¨OH£©₂]=0.3mol£¬¹Ê·´Ó¦ÖÐÏõËáÓÐÊ£Ó࣬
A£®CuÓëCu₂SµÄÎïÖʵÄÁ¿Ö®±È=0.25£º0.025=10£º1£¬¹ÊAÕýÈ·£»
B£®ÒÀ¾Ý¼ÆËã¿ÉÖª£¬·´Ó¦µÄÏõËáÉú³ÉÏõËáÍ­¡¢Ò»Ñõ»¯µª¡¢Ê£ÓàµÄÏõËᣬӦΪ1mol+0.2mol=1.2mol£¬Å¨¶È=$\frac{1.2mol}{0.5L}$=2.4mol/L£¬¹ÊB´íÎó£»
C£®Cu£¨OH£©₂³ÁµíÖÊÁ¿=98g/mol¡Á0.3mol=29.4g£¬¹ÊCÕýÈ·£»
D£®Cu¡¢CuSÓëÏõËá·´Ó¦ºóÊ£ÓàHNO₃=1mol+0.2mol-0.6mol-0.2mol=0.4mol£¬¹ÊDÕýÈ·£»
¹ÊÑ¡B£®

µãÆÀ ±¾Ì⿼²éÁËÑõ»¯»¹Ô­·´Ó¦µÄ¼ÆËã·ÖÎö£¬Ö÷ÒªÊÇÔªËØÊغ㡢µç×ÓÊغãµÄ¼ÆËãÓ¦Óã¬ÕÆÎÕ»ù´¡ÊǽâÌâ¹Ø¼ü£¬ÌâÄ¿ÄѶÈÖеȣ®