£¨1£©ÔÚ±ê×¼×´¿öÏ£¬COºÍCO2»ìºÏÆøÌåµÄÖÊÁ¿Îª36g£¬Ìå»ýΪ22.4L£¬ÔòCOËùÕ¼µÄÌå»ýΪ
11.2
11.2
L£¬ÖÊÁ¿Îª
14
14
g£¬»ìºÏÆøÌåÖÐCO2µÄ·Ö×ÓÊýΪ
0.5NA
0.5NA
£¬´Ë»ìºÍÆøÌåµÄƽ¾ùĦ¶ûÖÊÁ¿Îª
36g/mol
36g/mol
£®
£¨2£©3.01¡Á1023¸öOH-µÄÎïÖʵÄÁ¿Îª
0.5
0.5
mol£¬ÖÊÁ¿Îª
8.5
8.5
g£¬ÕâЩOH-Óë±ê×¼×´¿öÏÂÌå»ýΪ
11.2
11.2
LµÄNH3µÄÎïÖʵÄÁ¿ÏàµÈ£¬Óë
11.5
11.5
g Na+º¬ÓеÄÀë×ÓÊýÏàͬ£®
·ÖÎö£º£¨1£©¸ù¾Ýn=
V
Vm
¼ÆËã»ìºÏÆøÌåµÄÎïÖʵÄÁ¿£¬Áî»ìºÏÆøÌåÖÐCOºÍCO2µÄÎïÖʵÄÁ¿·Ö±ðΪxmol¡¢ymol£¬¸ù¾Ý¶þÕßÖÊÁ¿Ö®ºÍÓëÎïÖʵÄÁ¿Ö®ºÍÁзųö¼ÆËãx¡¢yµÄÖµ£¬ÔÙ¸ù¾ÝV=nVm¼ÆËãCOµÄÌå»ý£¬¸ù¾Ým=nM¼ÆËãCOµÄÖÊÁ¿£¬¸ù¾ÝN=nNA¼ÆËãCO2µÄ·Ö×ÓÊý£¬¸ù¾ÝM=
m
n
¼ÆËã»ìºÏÆøÌåµÄƽ¾ùĦ¶ûÖÊÁ¿£»
£¨2£©¸ù¾Ýn=
N
NA
¼ÆËãOH-µÄÎïÖʵÄÁ¿£¬¸ù¾Ým=nM¼ÆËãOH-µÄÖÊÁ¿£¬¸ù¾ÝV=nVm¼ÆËãNH3µÄÌå»ý£¬¸ù¾Ým=nM¼ÆËãNa+µÄÖÊÁ¿£®
½â´ð£º½â£º£¨1£©±ê×¼×´¿öÏ£¬»ìºÏÆøÌåµÄÎïÖʵÄÁ¿Îª
22.4L
22.4L/mol
=1mol£¬
Áî»ìºÏÆøÌåÖÐCOºÍCO2µÄÎïÖʵÄÁ¿·Ö±ðΪxmol¡¢ymol£¬Ôò£º
x+y=1
28x+44y=36
£¬½âµÃx=0.5£¬y=0.5£»
¹ÊCOµÄÌå»ýΪ0.5mol¡Á22.4L/mol=11.2L£¬ÖÊÁ¿Îª0.5mol¡Á28g/mol=14g£¬
CO2µÄ·Ö×ÓÊýΪ0.5mol¡ÁNAmol-1=0.5NA£¬
»ìºÏÆøÌåµÄƽ¾ùĦ¶ûÖÊÁ¿Îª
36g
1mol
=36g/mol£¬
¹Ê´ð°¸Îª£º11.2£»14£»0.5NA£»36g/mol£»
£¨2£©3.01¡Á1023¸öOH-µÄÎïÖʵÄÁ¿Îª
3.01¡Á1023
6.02¡Á1023mol-1
=0.5mol£¬ÖÊÁ¿Îª0.5mol¡Á17g/mol=8.5g£¬
ÓëOH-µÈÎïÖʵÄÁ¿µÄNH3µÄÌå»ýΪ0.5mol¡Á22.4L/mol=11.2L£¬ÓëOH-º¬ÓÐÏàͬµÄÀë×ÓÊýµÄNa+µÄÖÊÁ¿Îª0.5mol¡Á23g/mol=11.5g£¬
¹Ê´ð°¸Îª£º0.5£»8.5£»11.2£»11.5£®
µãÆÀ£º±¾Ì⿼²é»ìºÏÎïµÄÓйؼÆËã¡¢³£Óû¯Ñ§¼ÆÁ¿µÄÓйؼÆË㣬±È½Ï»ù´¡£¬×¢Òâ¶Ô¹«Ê½µÄÀí½âÕÆÎÕ£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

·ÖÎöijÖÖúÆøµÄÌå»ý×é³ÉÈçÏ£ºH2 50%¡¢CH4 30%¡¢CO 10%¡¢N2 6%¡¢CO2 4%£®
ÒÑÖª£ºH2£¨g£©+
1
2
O2£¨g£©¨TH2O£¨l£©¡÷H=-285.8kJ?mol-1
CO£¨g£©+
1
2
O2£¨g£©¨TCO2£¨g£©¡÷H=-282.6kJ?mol-1
CH4£¨g£©+2O2£¨g£©¨TCO2£¨g£©+2H2O£¨l£©¡÷H=-890.3kJ?mol-1
ÔòÔÚ±ê×¼×´¿öÏ£¬224L¸ÃÖÖúÆøȼÉÕʱ·Å³öµÄÈÈÁ¿Îª£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨1£©ÔÚ±ê×¼×´¿öÏ£¬ÆøÌåAµÄÃܶÈΪ1.25g/L£¬ÆøÌåBÏà¶ÔÓÚÇâÆøµÄÃܶÈΪ21£¬Èô8.96LAºÍBµÄ»ìºÏÆøÌåµÄÖÊÁ¿Îª13.44g£¬Ôò»ìºÏÎïÖÐAÓëBµÄÌå»ý±ÈΪ
3£º2
3£º2
£®
£¨2£©Í¬ÎÂͬѹϵÄSO2Ó뺤Æø£¬ÈôÖÊÁ¿Ïàͬ£¬Á½ÖÖÆøÌåµÄÌå»ý±ÈΪ
1£º16
1£º16
£»ÈôÌå»ýÏàͬʱ£¬Á½ÖÖÆøÌåµÄÖÊÁ¿±ÈΪ
16£º1
16£º1
£®
£¨3£©½«Ì¿·ÛÓëwgÑõÆøÖÃÓÚÃܱÕÈÝÆ÷ÖУ¬¾­¸ßγä·Ö·´Ó¦ºó£¬»Ö¸´µ½³õʼζȣ¬²âµÃ·´Ó¦Ç°ºóѹǿ·Ö±ðΪPo¡¢P£®ÒÑÖª£ºP=nPo£¬¼ÓÈëÌ¿·ÛÖÊÁ¿xgÓënµÄ±ä»¯¹ØϵÈçͼ£®
¢ÙnµÄ×îСֵΪ
1
1
£¬´ËʱxµÄÈ¡Öµ·¶Î§Îª
0£¼x¡Ü
3
8
w
0£¼x¡Ü
3
8
w
g£®¢ÚnµÄ×î´óֵΪ
2
2
£¬´ËʱxµÄÈ¡Öµ·¶Î§Îª
x¡Ý
3
4
w
x¡Ý
3
4
w
g£®¢ÛÈôʵÑé²âµÃnֵΪa£¬ÔòÈÝÆ÷ÄÚCO2ºÍCOµÄÌå»ý±ÈΪ
£¨2-a£©£º2£¨a-1£©
£¨2-a£©£º2£¨a-1£©
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨1£©ÔÚ±ê×¼×´¿öÏ£¬COºÍCO2»ìºÏÆøÌåµÄÖÊÁ¿Îª36g£¬Ìå»ýΪ22.4L£¬ÔòCOËùÕ¼µÄÌå»ýΪ
11.2
11.2
 L£¬ÖÊÁ¿Îª
14
14
g£¬»ìºÏÆøÌåÖÐCO2µÄ·Ö×ÓÊýΪ
0.5NA
0.5NA
£®
£¨2£©3.01¡Á1023¸öOH-µÄÎïÖʵÄÁ¿Îª
0.5mol
0.5mol
£¬ÖÊÁ¿Îª
8.5g
8.5g
£¬ÕâЩOH-Óë±ê×¼×´¿öÏÂÌå»ýΪ
11.2
11.2
LµÄNH3µÄÎïÖʵÄÁ¿ÏàµÈ£¬Óë
11.5
11.5
g Na+º¬ÓеÄÀë×ÓÊýÏàͬ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁйØÓÚÆøÌåĦ¶ûÌå»ýµÄ¼¸ÖÖ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢22.4 LÈκÎÆøÌåµÄÎïÖʵÄÁ¿¾ùΪ1 molB¡¢±ê×¼×´¿öÏ£¬1 molÎïÖʵÄÌå»ýΪ22.4 LC¡¢H2¡¢O2¡¢N2¡¢CO2×é³ÉµÄ»ìºÏÆøÌå1 molÔÚ±ê×¼×´¿öϵÄÌå»ýԼΪ22.4 LD¡¢ÔÚͬÎÂͬѹÏ£¬ÏàͬÌå»ýµÄÈκÎÆøÌåµ¥ÖÊËùº¬·Ö×ÓÊýºÍÔ­×ÓÊý¶¼Ïàͬ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸