ÌâÄ¿ÄÚÈÝ

ÎÀÉú²¿·¢³ö¹«¸æ£¬×Ô2011Äê5ÔÂ1ÈÕÆ𣬽ûÖ¹ÔÚÃæ·ÛÉú²úÖÐÌí¼Ó¹ýÑõ»¯¸Æ£¨CaO2£©µÈʳƷÌí¼Ó¼Á¡£¹ýÑõ»¯¸Æ£¨CaO2£©ÊÇÒ»ÖÖ°²È«ÎÞ¶¾ÎïÖÊ£¬´øÓнᾧˮ£¬Í¨³£»¹º¬ÓÐCaO¡£
£¨1£©³ÆÈ¡5.42 g¹ýÑõ»¯¸ÆÑùÆ·£¬×ÆÈÈʱ·¢ÉúÈçÏ·´Ó¦£º
2£ÛCaO2¡¤xH2O£Ý2CaO+O2¡ü+2xH2O£¬µÃµ½O2ÔÚ±ê×¼×´¿öÏÂÌå»ýΪ672 mL£¬¸ÃÑùÆ·ÖÐCaO2µÄÎïÖʵÄÁ¿Îª_____¡£
£¨2£©ÁíȡͬһÑùÆ·5.42 g£¬ÈÜÓÚÊÊÁ¿Ï¡ÑÎËáÖУ¬È»ºó¼ÓÈë×ãÁ¿µÄNa2CO3ÈÜÒº£¬½«ÈÜÒºÖÐCa2+È«²¿×ª»¯ÎªCaCO3³Áµí£¬µÃµ½¸ÉÔïµÄCaCO3 7.0 g¡£
¢ÙÑùÆ·ÖÐCaOµÄÖÊÁ¿Îª_____¡£
¢ÚÑùÆ·ÖÐCaO2¡¤xH2OµÄxֵΪ_____¡£
£¨1£© 0.06 mol  £¨2£©¢Ù0.56 g  ¢Ú1/2
£¨1£©n£¨CaO2£©=n£¨CaO2¡¤xH2O£©=2n£¨O2£©=2¡Á672 mL/22 400 mL¡¤mol-1=
0.06 mol¡£
£¨2£©¢Ùn£¨Ca2+£©×Ü=n£¨CaCO3£©="7.0" g¡Â100 g/mol="0.07" mol
m£¨CaO£©Ô­=£¨0.07 mol-0.06 mol£©¡Á56 g/mol="0.56" g
¢Úx=£¨5.42 g-0.56 g-0.06 mol¡Á72 g/mol£© ¡Â18 g/mol¡Â0.06 mol="1/2" ¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø