Question

6£®¼ºÖª£º²ÝËá¸Æ[CaC₂O₄]ÊÜÈȷֽ⣬CaC₂O₄$\frac{\underline{\;\;¡÷\;\;}}{\;}$CaCO₃+CO¡ü£®½«½«12.8g CaC₂O₄¹ÌÌå¼ÓÈÈÒ»¶Îʱ¼äºóÊ£Óà11.4g¹ÌÌ壨²»¿¼ÂÇ̼Ëá¸ÆµÄ·Ö½â£©£®ÏÂÁÐ˵·¨´íÎóµÄÕñ£¨¡¡¡¡£©

• A£® Ê£Óà¹ÌÌåΪ»ìºÏÎï
• B£® Ê£Óà¹ÌÌåÖк¬ÓиÆÔªËصÄÖÊÁ¿Îª4.0g
• C£® ½«Ê£Óà¹ÌÌåÈÜÓÚ¹ýÁ¿µÄÏ¡ÑÎËᣬÀíÂÛÉϲúÉú2.2g CO₂
• D£® ½«¼ÓÈȲúÉúµÄÆøÌåÈ«²¿È¼ÉտɲúÉú4.4g CO₂

Answer 1

·ÖÎö ¸ù¾Ý²ÝËá¸Æ¼ÓÈÈ»á·Ö½âÉú³É̼Ëá¸ÆºÍ¶þÑõ»¯Ì¼£¬ÒÀ¾ÝÌâÖеÄÊý¾ÝºÍÑ¡Ïî½øÐмÆËã·ÖÎö£®

½â´ð ½â£ºÉú³ÉÒ»Ñõ»¯Ì¼ÖÊÁ¿Îª12.8g-11.4g=1.4g£¬
ÉèÏûºÄCaC₂O₄ÖÊÁ¿Îªx£¬µÃµ½Ì¼Ëá¸ÆµÄÖÊÁ¿Îªy
CaC₂O₄$\frac{\underline{\;\;¡÷\;\;}}{\;}$CaCO₃+CO¡ü
128              100      28
x                   y       1.4g
$\frac{128}{x}=\frac{100}{y}=\frac{28}{1.4g}$
x=6.4g
y=5g
·Ö½âµÄ̼Ëá¸ÆÓë¹ýÁ¿µÄÏ¡ÑÎËá·´Ó¦²úÉúCO₂ÖÊÁ¿Îªz
CaCO₃ +2HCl=CaCl₂+CO₂¡ü+H₂O
 100                           44
   5g                           z
$\frac{100}{5g}=\frac{44}{z}$
z=2.2g
²úÉúµÄÆøÌåÈ«²¿È¼ÉտɲúÉúCO₂ÖÊÁ¿Îªa
2CO+O₂$\frac{\underline{\;µãȼ\;}}{\;}$2CO₂£®
 56                  88
 1.4g                a
$\frac{56}{1.4g}=\frac{88}{a}$     
a=2.2g  
A¡¢ÓɼÆËã½á¹û¿ÉÖª£¬Ê£Óà¹ÌÌåΪ²ÝËá¸ÆºÍ̼Ëá¸ÆµÄ»ìºÏÎ¹ÊAÕýÈ·£»
B¡¢¸ÆÔªËØÖÊÁ¿Êغ㣬ËùÒÔÊ£Óà¹ÌÌåÖк¬ÓиÆÔªËصÄÖÊÁ¿Îª12.8¡Á$\frac{40}{128}$¡Á100%=4.0g£¬¹ÊBÕýÈ·£»
C¡¢Ê£Óà̼Ëá¸ÆµÄÖÊÁ¿Îª5g£¬ÈÜÓÚ¹ýÁ¿µÄÏ¡ÑÎËᣬ²úÉú2.2g¶þÑõ»¯Ì¼£¬¹ÊCÕýÈ·£»
D¡¢Í¨¹ý·ÖÎö¿ÉÖª£¬½«¼ÓÈȲúÉúµÄÆøÌåÈ«²¿È¼ÉտɲúÉú2.2g CO₂£¬¹ÊD´íÎó£®
¹ÊÑ¡£ºD£®

µãÆÀ ±¾ÌâÊôÓÚʵÑé̽¾¿Ì⣬×ÛºÏÐÔ¾¿£¬Ê×ÏÈ·ÖÎöÑ¡ÏȻºó½áºÏÌâÖеķ´Ó¦Ô­Àí½øÐмÆË㣬ֻÓÐ×ÛºÏÕÆÎÕÁË»ù´¡ÖªÊ¶²ÅÄܽϺõÄÍê³É±¾ÀàÁ·Ï°Ì⣮