ÌâÄ¿ÄÚÈÝ
15£®»ØÊÕǦÐîµç³ØµÄµç¼«Ìî³äÎǦ¸à£¬Ö÷Òªº¬PbO¡¢PbO2¡¢PbSO4£©£¬¿ÉÖƱ¸ÈÈÎȶ¨¼ÁÈýÑλùÁòËáǦ£¨×é³É¿É±íʾΪ3PbO•PbSO4•H2O£©£¬ÆäʵÑéÁ÷³ÌÈçͼ£º£¨1£©ÎïÖÊX¿ÉÒÔÑ»·ÀûÓ㬸ÃÎïÖÊÊÇÏõËᣮ¼ìÑéÁ÷³ÌÖÐʹÓõÄNa2SO3ÈÜÒºÊÇ·ñ±äÖʵķ½·¨ÊÇÈ¡ÉÙÁ¿ÈÜÒº£¬¼ÓÈë×ãÁ¿Ï¡ÑÎËᣬÔÙ¼ÓÂÈ»¯±µÈÜÒº£¬¹Û²ìÊÇ·ñ²úÉú°×É«³Áµí£®
£¨2£©´ÓÂËÒºA¿ÉÌáÈ¡³öÒ»ÖÖº¬½á¾§Ë®µÄÄÆÑθ±²úÆ·£®Èô¼ìÑé¸Ã¾§ÌåÖнᾧˮµÄº¬Á¿£¬ËùÐèµÄ¹èËáÑβÄÖÊÒÇÆ÷ÓÐÛáÛö¡¢¾Æ¾«µÆ¡¢ÄàÈý½Ç¡¢²£Á§°ô¡¢Ê¯ÃÞÍø¡¢¸ÉÔïÆ÷µÈ£®
£¨3£©Á÷³ÌÖв»Ö±½ÓÀûÓÃH2SO4ÈÜÒºÓëPbO¡¢PbCO3·´Ó¦ÖÆÈ¡PbSO4£¬ÔÒò¿ÉÄÜÊÇPbSO4²»ÈÜÓÚË®£¬¸²¸ÇÔÚ¹ÌÌå±íÃæ×è°·´Ó¦µÄ½øÒ»²½·¢Éú£®
£¨4£©Éú³ÉÈýÑλùÁòËáǦµÄÀë×Ó·´Ó¦·½³ÌʽΪ4PbSO4+6OH-=3PbO•PbSO4•H2O+3SO42-+2H2O£®
£¨5£©ÏòǦ¸à½¬ÒºÖмÓÈëNa2SO3ÈÜÒºµÄÄ¿µÄÊǽ«ÆäÖеÄPbO2»¹ÔΪPbO£®ÈôʵÑéÖÐËùȡǦ¸àµÄÖÊÁ¿Îª47.8g£¬ÆäÖÐPbO2µÄÖÊÁ¿·ÖÊýΪ15.0%£¬ÔòÒª½«PbO2È«²¿»¹Ô£¬ÖÁÉÙÐè¼Ó30mL 1.0mol•L-1 Na2SO3ÈÜÒº£®
·ÖÎö £¨1£©ÀûÓ÷ϾÉǦÐîµç³ØÒõ¡¢Ñô¼«Ìî³äÎǦ¸à£©ÖƱ¸ËÜÁϼӹ¤ÈÈÎȶ¨¼ÁÈýÑλùÁòËáǦʵÑéÁ÷³Ì£®ÏòǦ¸à½¬ÒºÖмÓÈëNa2SO3ÈÜÒºµÄÄ¿µÄÊǽ«PbO2»¹ÔPbO£¬Na2SO3+PbO2=PbO+Na2SO4£¬¼ÓNa2CO3ÈÜÒºÊǽ«PbSO4ת»¯³ÉPbCO3£¬ËùÒÔÂËÒº¢ñÖ÷ÒªÊÇNa2SO4ÈÜÒº£®PbOºÍPbCO3ÔÚÏõËáµÄ×÷ÓÃÏÂת»¯³ÉPb£¨NO3£©£®Pb£¨NO3£©ÖмÓÏ¡H2SO4ת»¯³ÉPbSO4ºÍÏõËᣬÒò´ËXΪHNO3£¬¿ÉÑ»·ÀûÓã»Èç¹û±äÖÊ˵Ã÷±ä³ÉÁòËá±µ£¬¿ÉÒÔͨ¹ý¼ìÑéÊÇ·ñÉú³ÉÁòËá¸ùÀë×Ó£»
£¨2£©²âÁ¿¾§ÌåÖнᾧˮµÄº¬Á¿£¬ÊµÑé²½ÖèΪ£º¢ÙÑÐÄ¥ ¢Ú³ÆÁ¿¿ÕÛáÛöºÍ×°ÓÐÊÔÑùµÄÛáÛöµÄÖÊÁ¿ ¢Û¼ÓÈÈ ¢ÜÀäÈ´ ¢Ý³ÆÁ¿ ¢ÞÖظ´¢ÛÖÁ¢ÝµÄ²Ù×÷£¬Ö±µ½Á¬ÐøÁ½´Î³ÆÁ¿µÄÖÊÁ¿²î²»³¬¹ý0.1gΪֹ ¢ß¸ù¾ÝʵÑéÊý¾Ý¼ÆË㾧ÌåÖнᾧˮµÄº¬Á¿£¬ËùÒÔËùÐèµÄ¹èËáÑβÄÖÊÒÇÆ÷Óв£Á§°ô¡¢Ê¯ÃÞÍø¡¢¸ÉÔïÆ÷¡¢ÛáÛö¡¢¾Æ¾«µÆ¡¢ÄàÈý½Ç£»
£¨3£©Éú³ÉµÄÁòËáǦÄÑÈÜÓÚË®£¬¸²¸ÇÔÚ¹ÌÌåPbO¡¢PbCO3µÄ±íÃ棬×è°·´Ó¦µÄ½øÒ»²½·¢Éú£»
£¨4£©´ÓÁ÷³Ì¿´£¬ÁòËáǦºÍÇâÑõ»¯ÄÆ·´Ó¦Éú³ÉÈýÑλùÁòËáǦºÍÁòËáÄÆ£¬¾Ý´Ëд³ö·½³Ìʽ£»
£¨5£©¸ù¾ÝǦ¸àµÄÖÊÁ¿¼°Ñõ»¯Ç¦µÄÖÊÁ¿·ÖÊý¿É¼ÆËã³öÑõ»¯Ç¦µÄÎïÖʵÄÁ¿£¬¸ù¾ÝÑÇÁòËáÄƺÍÑõ»¯Ç¦·´Ó¦¹Øϵ¿É¼ÆËãÐèÒªÑÇÁòËáÄƵÄÌå»ý£®
½â´ð ½â£º£¨1£©·ÖÎöÁ÷³Ì¿ÉÖª£¬PbOºÍPbCO3ÔÚÏõËáµÄ×÷ÓÃÏÂת»¯³ÉPb£¨NO3£©£®Pb£¨NO3£©ÖмÓÏ¡H2SO4ת»¯³ÉPbSO4ºÍÏõËᣬÒò´ËXΪHNO3£¬¿ÉÑ»·ÀûÓ㬼ìÑéÁòËá¸ùÀë×ӵķ½·¨Îª£ºÈ¡ÉÙÁ¿¾§ÌåÈÜÓÚÕôÁóË®£¬È»ºóÓÃÑÎËáËữ£¬ÔÙµÎBaCl2ÈÜÒº£¬Èô³öÏÖ°×É«³Áµí£¬¼´Ö¤Ã÷¸Ã¾§ÌåÖк¬ÓÐSO42-£¬
¹Ê´ð°¸Îª£ºÏõË᣻ȡÉÙÁ¿ÈÜÒº£¬¼ÓÈë×ãÁ¿Ï¡ÑÎËᣬÔÙ¼ÓÂÈ»¯±µÈÜÒº£¬¹Û²ìÊÇ·ñ²úÉú°×É«³Áµí£»
£¨2£©²âÁ¿¾§ÌåÖнᾧˮµÄº¬Á¿£¬ÊµÑé²½ÖèΪ£º¢ÙÑÐÄ¥ ¢Ú³ÆÁ¿¿ÕÛáÛöºÍ×°ÓÐÊÔÑùµÄÛáÛöµÄÖÊÁ¿ ¢Û¼ÓÈÈ ¢ÜÀäÈ´ ¢Ý³ÆÁ¿ ¢ÞÖظ´¢ÛÖÁ¢ÝµÄ²Ù×÷£¬Ö±µ½Á¬ÐøÁ½´Î³ÆÁ¿µÄÖÊÁ¿²î²»³¬¹ý0.1gΪֹ ¢ß¸ù¾ÝʵÑéÊý¾Ý¼ÆË㾧ÌåÖнᾧˮµÄº¬Á¿£¬ËùÒÔËùÐèµÄ¹èËáÑβÄÖÊÒÇÆ÷Óв£Á§°ô¡¢Ê¯ÃÞÍø¡¢¸ÉÔïÆ÷¡¢ÛáÛö¡¢¾Æ¾«µÆ¡¢ÄàÈý½Ç£¬
¹Ê´ð°¸Îª£ºÛáÛö£»¾Æ¾«µÆ£»ÄàÈý½Ç£»
£¨3£©Éú³ÉµÄÁòËáǦÄÑÈÜÓÚË®£¬¸²¸ÇÔÚ¹ÌÌåPbO¡¢PbCO3µÄ±íÃ棬×è°·´Ó¦µÄ½øÒ»²½·¢Éú£¬¹Ê´ð°¸Îª£ºPbSO4²»ÈÜÓÚË®£¬¸²¸ÇÔÚ¹ÌÌå±íÃæ×è°·´Ó¦µÄ½øÒ»²½·¢Éú£»
£¨4£©´ÓÁ÷³Ì¿´£¬ÁòËáǦºÍÇâÑõ»¯ÄÆ·´Ó¦Éú³ÉÈýÑλùÁòËáǦºÍÁòËáÄÆ£¬·´Ó¦·½³ÌʽΪ£º4PbSO4+6NaOH=3PbO•PbSO4•H2O+3Na2SO4+2H2O£¬Àë×Ó·½³ÌʽΪ£º
4PbSO4+6OH-=3PbO•PbSO4•H2O+3SO42-+2H2O£¬¹Ê´ð°¸Îª£º4PbSO4+6OH-=3PbO•PbSO4•H2O+3SO42-+2H2O£»
£¨5£©Ñõ»¯Ç¦µÄÎïÖʵÄÁ¿Îª£º$\frac{47.8g¡Á15%}{239g/mol}$=0.03mol£¬
PbO2 ¡«Na2SO3
1mol 1mol
0.03mol n
n=0.03mol£¬V=$\frac{0.03mol}{1.0mol/L}$=0.03L=30mL£¬
¹Ê´ð°¸Îª£º30£®
µãÆÀ ±¾Ì⿼²éÁËʵÑé·½°¸µÄÉè¼Æ£¬ÖеÈÄѶȣ®Òª·ÖÎöÁ÷³Ì£¬´ÓÁ÷³ÌÖÐËù¸øÐÅÏ¢½áºÏÌâÄ¿ÉèÎʽâÌ⣮
£¨1£©ZµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïWÊÇÖØÒªµÄ»¯¹¤ÔÁÏ£®WµÄ»¯Ñ§Ê½ÎªH2SO4£»¹¤ÒµÖÆÔìWµÄÉú²ú¹ý³ÌÖ÷Òª·ÖΪÈý¸ö½×¶Î£®
¢Ù101kPaʱ£¬3.2g ZµÄ¹ÌÌåµ¥ÖÊÍêȫȼÉտɷųö29.7kJµÄÈÈÁ¿£¬Ð´³öÄܹ»±íʾ¸Ã¹ÌÌåµ¥ÖÊȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽS£¨s£©+O2£¨g£©=SO2£¨g£©¡÷H=-297kJ/mol£»
¢ÚÔÚ½Ó´¥Ñõ»¯½×¶Î£¬ÎªÌá¸ßZY2µÄת»¯ÂÊ£¬´ÓÀíÂÛÉÏÅжϣ¬Ó¦Ñ¡ÔñµÄÌõ¼þÊǵÍκ͸ßѹ£®´Ó±í¸ø³ö²»Í¬Î¶ȡ¢Ñ¹Ç¿ÏÂZY2ƽºâת»¯ÂʵÄʵÑéÊý¾Ý½øÐзÖÎö£¬½áºÏ¹¤ÒµÉú²úµÄʵ¼Ê£¬Ó¦Ñ¡ÔñÊʺϵÄζȺÍѹǿÊÇ£¨Ñ¡Ìî×Öĸ£©B£»
A£®400¡æ¡«500¡æ10MPa B£®400¡æ¡«500¡æ1MPa
C£®500¡æ¡«500¡æ10MPa D£®400¡æ¡«500¡æ0.1MPa
ѹǿ/MPa ת»¯ÂÊ/% ζÈ/¡æ | 0.1 | 0.5 | 1 | 10 |
400 | 99.2 | 99.6 | 99.7 | 99.9 |
500 | 93.5 | 96.9 | 97.8 | 99.3 |
600 | 73.7 | 85.8 | 89.5 | 96.4 |
A£®Ë® B£®0.5mol/LµÄÁòËá C£®98.3%µÄÁòËá D£®Å¨°±Ë®
£¨2£©ÒÑÖªXÓëXY¶¼Êǹ¤ÒµÉϳ£ÓõĻ¹Ô¼Á£®
¢Ùд³öXµ¥ÖÊÓëWµÄŨÈÜÒº·´Ó¦µÄ»¯Ñ§·½³ÌʽC+2H2SO4£¨Å¨£©$\frac{\underline{\;\;¡÷\;\;}}{\;}$CO2¡ü+2SO2¡ü+2H2O£»
¢Ú500¡æ£¬11.2L£¨±ê×¼×´¿ö£©ZY2ÔÚ´ß»¯¼Á×÷ÓÃÏÂÓëXY·¢Éú»¯Ñ§·´Ó¦£®ÈôÓÐ2¡Á6.02¡Á1023¸öµç×ÓתÒÆʱ£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇSO2+2CO$\frac{\underline{´ß»¯¼Á}}{¡÷}$S+2CO2£®
A£® | SO2ÊÇÑõ»¯²úÎï | |
B£® | CuFeS2½ö×÷»¹Ô¼Á£¬ÁòÔªËر»Ñõ»¯ | |
C£® | ÿÉú³É1mol Cu2S£¬ÓÐ4molÁò±»Ñõ»¯ | |
D£® | ÿתÒÆ1.2molµç×Ó£¬ÓÐ0.2molÁò±»Ñõ»¯ |
£¨1£©Ê³Æ·Ìí¼Ó¼Áï§Ã÷·¯NH4Al£¨SO4£©2•12H2O¸ßοɷֽ⣬ÏÂÁйØÓÚÆä·Ö½â²úÎïµÄÔ¤²â²»ºÏÀí µÄÊÇC£®
A£®NH3¡¢N2¡¢SO2¡¢H2O B£®NH3¡¢SO3¡¢H2O
C£®NH3¡¢SO2¡¢H2O D£®NH3¡¢N2¡¢SO3¡¢SO2¡¢H2O
£¨2£©Æû³µ·¢¶¯»ú¹¤×÷ʱҲ»áÒý·¢N2ºÍO2·´Ó¦²úÉú´óÆøÎÛȾÎïNO£¬ÆäÄÜÁ¿±ä»¯Ê¾ÒâͼÈçͼ1Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪN2£¨g£©+O2£¨g£©=2NO£¨g£©¡÷H=+183.8kJ/mol£®
£¨3£©¹¤ÒµºÏ³É°±µÄ·´Ó¦£ºN2£¨g£©+3H2£¨g£©?2NH3£¨g£©¡÷H£¼0£¬·Ö±ðÔÚT1¡¢T2ζÈÏ£¬¸Ä±äÆðʼÇâÆøÎïÖʵÄÁ¿£¬²âµÃƽºâʱ°±µÄÌå»ý·ÖÊýÈçͼ2Ëùʾ£º
¢Ù±È½ÏÔÚm¡¢n¡¢qÈýµãËù´¦µÄƽºâ״̬ÖУ¬·´Ó¦ÎïN2µÄת»¯ÂÊ×î¸ßµÄÊÇPµã£®
¢ÚT2Ìõ¼þÏ£¬ÔÚ2LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈëx mol N2ºÍ y mol H2ʱ£¬3min´ïƽºâ£¬´Ëʱ·´Ó¦ÎïµÄת»¯ÂʾùΪa£¬Ð´³öÏÂÁнöº¬a¡¢xµÄ±í´ïʽ£¨²»±Ø»¯¼ò£©£ºv£¨N2£©=$\frac{xa}{6}$mol•L-1•min-1£»¸Ã·´Ó¦µÄƽºâ³£ÊýµÄÖµK=$\frac{16£¨xa£©{\;}^{2}}{£¨x-xa£©£¨3x-3xa£©{\;}^{3}}$£®
¢ÛͼÏóÖÐT2µÍÓÚT1£¨Ìî¡°¸ßÓÚ¡±¡¢¡°µÍÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°ÎÞ·¨È·¶¨¡±£©£®
¢Ü¿Æѧ¼Ò²ÉÓøßÖÊ×Óµ¼µçÐÔµÄSCYÌÕ´É£¨ÄÜ´«µÝH+ £©ÊµÏÖ°±µÄµç»¯Ñ§ºÏ³É£¬ÕâÌá¸ßÁ˵ªÆøºÍÇâÆøµÄת»¯ÂÊ£®Ð´³öµç»¯Ñ§ºÏ³É¹ý³ÌÖз¢Éú»¹Ô·´Ó¦µÄµç¼«·½³Ìʽ£ºN2+6H++6e-=2NH3£®
£¨4£©½«ÖÊÁ¿ÏàµÈµÄËÄ·ÝÌú·ÛºÍͷ۵ľùÔÈ»ìºÏÎ·Ö±ð¼ÓÈëͬŨ¶ÈÏ¡ÏõËá³ä·Ö·´Ó¦£¬£¨¼ÙÉèÏõËáµÄ»¹Ô²úÎïÖ»ÓÐNO£©ÊµÑéÊý¾ÝÈçÏÂ±í£º
±à ºÅ | ¢Ù | ¢Ú | ¢Û | ¢Ü |
Ï¡ÏõËáÌå»ý/mL | 100mL | 200mL | 300mL | 400mL |
Ê£Óà½ðÊô/g | 18.0g | 9.6g | 0 | 0 |
NOÌå»ý/L£¨±ê×¼×´¿öÏ£© | 2.24L | 4.48L | 6.72L | V |
A£®ÏõËáÆðʼŨ¶ÈΪ4mol/L B£®¢ÙÖÐÈܽâÁË5.6g Fe
C£®¢ÛÖÐn£¨Cu2+£©=0.15mol D£®¢ÜÖÐV=6.72L
A£® | ÓÍ֬ˮ½â¿ÉµÃµ½°±»ùËáºÍ¸ÊÓÍ | |
B£® | ·À¸¯¼Á¶¼ÓÐÒ»¶¨µÄ¶¾ÐÔ£¬ËùÒÔ²»ÄܳԺ¬·À¸¯¼ÁµÄʳƷ | |
C£® | µ°°×ÖÊÈÜÒº¡¢µí·ÛÈÜÒº¶¼Êô½ºÌå | |
D£® | ²£Á§¸ÖÊÇÐÂÐ͵ÄÎÞ»ú·Ç½ðÊô²ÄÁÏ |
A£® | ¹èËáÄÆÈÜÒºÓë´×ËáÈÜÒº»ìºÏ£ºSiO32-+2H+=H2SiO3¡ý | |
B£® | NH4Al£¨SO4£©2ÈÜÒºÓë¹ýÁ¿Ï¡°±Ë®·´Ó¦£ºAl3++3NH3•H2O=Al£¨OH£©3¡ý+3NH4+ | |
C£® | ÓÃÏ¡ÏõËáÇåÏ´ÊÔ¹ÜÄÚ±ÚµÄÒø¾µ£ºAg+2H++NO3-=Ag++NO2¡ü+H2O | |
D£® | FeBr2ÈÜÒºÖÐͨÈë¹ýÁ¿Cl2£º2Fe2++2Br-+2Cl2=2Fe3++Br2+4Cl- |