在电解饱和食盐水的过程中，当阴阳两极共收集到224mL气体(标准状况)时， 理论上可得到氢氧化钠 ( )g
\u5728\u7535\u89e3\u9971\u548c\u98df\u76d0\u6c34\u7684\u8fc7\u7a0b\u4e2d\uff0c\u5f53\u9634\u9633\u4e24\u6781\u5171\u6536\u96c6\u5230224mL\u6c14\u4f53(\u6807\u51c6\u72b6\u51b5)\u65f6\uff0c \u7406\u8bba\u4e0a\u53ef\u5f97\u5230\u6c22\u6c27\u5316\u94a0 ( )g0.4g
\u6c14\u4f53\u7684\u7269\u8d28\u7684\u91cf\u662fn=V/Vm=0.224/22.4=0.01mol
2NaCl+2H2O=\u901a\u7535=2NaOH+H2\u2191+Cl2\u2191
\u901a\u8fc7\u65b9\u7a0b\u5f0f\u53ef\u4ee5\u770b\u51fa\uff0c\u751f\u6210\u7684H2\u4e0eCl2\u7684\u7269\u8d28\u7684\u91cf\u662f\u76f8\u7b49\u7684\uff0c\u6240\u4ee5H2\u548cCl2\u7684\u7269\u8d28\u7684\u91cf\u90fd\u662f0.01/2=0.005mol
2NaCl+2H2O=\u901a\u7535=2NaOH+H2\u2191+Cl2\u2191
2 1
n 0.005mol
n=0.01mol
NaOH\u7684\u8d28\u91cf\u662fm=nM=0.01*40=0.4g
1\u3001\u9632\u6b62\u6c2f\u6c14\u548c\u6c22\u6c27\u5316\u94a0\u53cd\u5e94
2\u3001\u9632\u6b62\u6c2f\u6c14\u548c\u6c22\u6c14\u53cd\u5e94
2NaCl+2H2O=通电=2NaOH+H2↑+Cl2↑
生成的氢气和氯气的物质的量相同，总量是n=0.224/22.4=0.01mol，所以氢气和氯气的物质的量=n/2=0.01/2=0.05mol
是224ml化成0.224l再计算物质的量得0.224/22.4=0.01mol
阴阳两极共收集到224mL气体，而且是1:1的关系，所以分半要除以2.
这样才能用NaOH：H2=2:1来算
绛旓細X 0.005 X= 0.4g
绛旓細闃撮槼涓ゆ瀬鍏辨敹闆嗗埌224mL姘斾綋 锛屾皵浣撶殑鐗╄川鐨勯噺鏄0.01mol 杩欐槸鎬荤殑锛孒2 Cl2鍚勫崰涓鍗婏紝鎵浠ラ兘闄や互2
绛旓細0.4g 姘斾綋鐨勭墿璐ㄧ殑閲忔槸n=V/Vm=0.224/22.4=0.01mol 2NaCl+2H2O=閫氱數=2NaOH+H2鈫+Cl2鈫 閫氳繃鏂圭▼寮忓彲浠ョ湅鍑猴紝鐢熸垚鐨凥2涓嶤l2鐨勭墿璐ㄧ殑閲忔槸鐩哥瓑鐨勶紝鎵浠2鍜孋l2鐨勭墿璐ㄧ殑閲忛兘鏄0.01/2=0.005mol 2NaCl+2H2O=閫氱數=2NaOH+H2鈫+Cl2鈫 2 1 n 0.005mol n=0.01mol NaOH鐨勮川閲忔槸m=...
绛旓細鐢佃В楗卞拰椋熺洂姘锛岄槼鏋佸彂鐢2Cl - -2e - =Cl 2 鈫戯紝闃存瀬鍙戠敓4H 2 O+4e - =2H 2 鈫+4OH - 锛岀敱鐢靛瓙瀹堟亽鍙煡锛2Cl 2 鈫戯綖4OH - 锛岄槾鏋侀檮杩戞湁0.4molOH - 鐢熸垚鏃讹紝闃虫瀬鐢熸垚姘皵涓 0.4mol 2 =0.2mol锛屾晠閫塀锛
绛旓細鐢佃В楗卞拰椋熺洂姘锛岄槾鏋佷笂鐨勭數鏋佸弽搴斿紡涓2H++2e-=H2鈫戯紝褰撻槾鏋侀檮杩戞湁0.4molOH-鐢熸垚鏃讹紝璇存槑鍦ㄨ鏋佷笂鍑忓皯鐨勬阿绂诲瓙涓0.4mol锛屾墍浠ヤ細杞Щ鐢靛瓙0.4mol锛岄槼鏋佺數鏋佸弽搴斿紡涓2Cl--2e-鈺怌l2锛岀敓鎴0.2molCl2锛庢晠閫塀锛
绛旓細鐢佃В楗卞拰椋熺洂姘锛岄槾鏋佷笂鐨勭數鏋佸弽搴斿紡涓2H++2e-鈺怘2鈫锛屽綋闃鏋侀檮杩戞湁0.8mol NaOH鐢熸垚锛屽嵆0.8mol OH-鐢熸垚鏃讹紝璇存槑鍦ㄨ鏋佷笂鍑忓皯鐨勬阿绂诲瓙涓0.8mol锛屾墍浠ヤ細杞Щ鐢靛瓙0.8mol锛庢晠閫夛細B锛
绛旓細鐢佃В楗卞拰椋熺洂姘寸殑杩囩▼涓,闃存瀬鍙戠敓杩樺師鍙嶅簲,姘㈢瀛愬緱鐢靛瓙,鐢熸垚姘㈡皵.鑰屾阿绂诲瓙鏄按鐢电鍑虹殑,鍏跺弽搴斿悗鍓╀綑鐨勪负姘㈡哀鏍圭瀛愶紝鍥犳锛岄槾鏋佷骇鐢烵H-銆
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