实验室用10g大理石与足量稀盐酸反应.(1)至少需要2mol /l 稀盐酸多少毫升
设需要HCl xmolCaCO3+2HCl=CaCl2+H2O+CO2
100 2
10 x
100/10=2/x 解得x=0.2mol
根据量的浓度=溶质物质的量/ 溶液体积
2molL-=0.2mol / V
V=0.1升
绛旓細鐢熸垚鐨勪簩姘у寲纰抽氬叆婢勬竻鐭崇伆姘达紝鍙戠幇鐭崇伆姘村閲4 .4g,鍗充簩姘у寲纰崇殑璐ㄩ噺涓4.4g CaCO3+2HCl=CaCl2+H2O+CO2 100 44 x 4.4g 100/x=44/4.4g x=10g 10g/20g*100%=50
绛旓細鐢熸垚0.1molCO2鑷冲皯闇瑕0.1mol纰抽吀閽欙紝鍗10g銆傜敓鎴愮殑CO2涓嶩2O鐗╄川閲忔瘮涓1锛1锛屾墍浠ユ按涔熸槸0.1mol 楂樹腑棰橀毦锛屽鍋氬仛灏卞ソ浜嗭紝鍗充娇鍐嶇畝鍗曠殑棰橈紝涓嶄細涔熻闂紝杩欐牱瀛︿範鎴愮哗鎵嶈兘杩涙 甯屾湜瀵逛綘鐨勫涔犳湁甯姪鍝︼紝涓嶆噦鍐嶉棶鍚
绛旓細锛1锛変骇鐢熶簩姘у寲纰崇殑璐ㄩ噺涓猴細100g+10g-107.8g=2.2g锛涳紙2锛夎璇澶х悊鐭涓⒊閰搁挋鐨勮川閲忎负xCaCO3+2HCl鈺怌aCl2+CO2鈫+H2O100 44x 2.2g10044锛漻2.2gx=5g澶х悊鐭充腑纰抽吀閽欑殑璐ㄩ噺鍒嗘暟锛5g10g脳100%=50%锛庣瓟锛氾紙1锛変骇鐢熶簩姘у寲纰崇殑璐ㄩ噺涓2.2g锛涳紙2锛夊ぇ鐞嗙煶涓⒊閰搁挋鐨勮川閲忓垎鏁颁负50%锛
绛旓細锛1锛5锛2锛绋鐩閰稿厛涓庣⒊閰搁挔鍙嶅簲銆傦紙3锛夎В锛氳鍘熸护娑蹭腑CaCl 2 鐨勮川閲忎负xNa 2 CO 3 +CaCl 2 = CaCO 3 鈫+ 2NaCl 111 100 x 5g x= 鍘熸护娑蹭腑CaCl 2 鐨勮川閲忓垎鏁= 绛旓細鍘熸护娑蹭腑姘寲閽欑殑璐ㄩ噺鍒嗘暟鏄11.1%銆
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