一道高一的化学计量计算题
\u4e00\u9053\u9ad8\u4e00\u7684\u5316\u5b66\u8ba1\u91cf\u8ba1\u7b97\u9898\uff0c\u6c42\u5e2e\u5fd9\uff01\u6c2f\u5316\u94a0\u5206\u5b50\u5f0f\uff08NaCl\uff09\u4e00\u4e2a\u6c2f\u5316\u94a0\u5206\u5b50\u5728\u6eb6\u6db2\u4e2d\u53ef\u4ee5\u7535\u79bb\u51fa\u4e00\u4e2a\u94a0\u79bb\u5b50\u548c\u4e00\u4e2a\u6c2f\u79bb\u5b50\uff0c\u6240\u4ee5\u6d53\u5ea6\u662f\u4e00\u6837\u7684\u3002
100ML,0.10mol/L\u7684\u6c2f\u5316\u94a0\u6eb6\u6db2\u6240\u542b\u7684 \u6c2f\u79bb\u5b50\u4e3a0.10mol/L
A\uff1a100ML,0.5mol/L\u7684\u6c2f\u5316\u9541\u6eb6\u6db2 \uff08MgCl2)\u6c2f\u79bb\u5b50\u4e3a \u6c2f\u5316\u9541\u7684\u4e24\u500d 1mol/L
B\uff1a200ML,0.25mol/L\u7684\u6c2f\u5316\u9499\u6eb6\u6db2 \uff08CaCl2)\u6c2f\u79bb\u5b50\u4e3a \u6c2f\u5316\u9499\u7684\u4e24\u500d 0.5mol/L
C:50ML,1mol/L\u7684\u6c2f\u5316\u94a0\u6eb6\u6db2 \u6c2f\u79bb\u5b50\u4e3a1mol/L
D\uff1a25ML,0.5mol/L\u7684\u6c2f\u5316\u94dd\u6eb6\u6db2 \uff08AlCl3)\u6c2f\u79bb\u5b50\u4e3a 1.5mol/L
\u8fd9\u4e2a\u9898\u76ee\u4f60\u662f\u4e0d\u662f\u770b\u9519\u4e86 \u7269\u8d28\u7684\u91cf
100ML,0.10mol/L\u7684\u6c2f\u5316\u94a0\u6eb6\u6db2 \u7684\u662f0.01mol
A\uff1a100ML,0.5mol/L\u7684\u6c2f\u5316\u9541\u6eb6\u6db2 \uff08MgCl2)\u6c2f\u79bb\u5b50\u4e3a 0.1mol
B\uff1a200ML,0.25mol/L\u7684\u6c2f\u5316\u9499\u6eb6\u6db2 \uff08CaCl2)\u6c2f\u79bb\u5b50\u4e3a 0.1mol
C:50ML,1mol/L\u7684\u6c2f\u5316\u94a0\u6eb6\u6db2 \u6c2f\u79bb\u5b50\u4e3a0.05mol
D\uff1a25ML,0.5mol/L\u7684\u6c2f\u5316\u94dd\u6eb6\u6db2 \uff08AlCl3)\u6c2f\u79bb\u5b50\u4e3a0.0375mol
\u8c8c\u4f3c\u6ca1\u6709\u7b54\u6848
Mg\u548cNaOH\u4e0d\u53cd\u5e94
\u800c2Al+2NaOH+2H2O=2NaAlO2+3H2\u2191
\u90a3\u4e48Mg\u7684\u8d28\u91cf\u662fcg
Al\u7684\u8d28\u91cf\u662fa-cg
\u800c\u751f\u6210\u7684\u6c22\u6c14\u4f53\u79ef\u662fbmL=b/1000L
\u90a3\u4e48\u7269\u8d28\u7684\u91cf\u662f(b/1000)/22.4=b/22400mol
\u8bbeAl\u7684\u76f8\u5bf9\u539f\u5b50\u91cf\u662fM
\u5219Al\u7684\u7269\u8d28\u7684\u91cf\u662f\uff08a-c\uff09/M
\u6839\u636e\u6bd4\u4f8b\u5f0f\u77e5\u9053\u4e86Al\u7684\u7269\u8d28\u7684\u91cf\u548cH2\u7684\u7269\u8d28\u7684\u91cf\u7684\u5173\u7cfb\u662f
2:3=\uff08a-c\uff09/M\uff1ab/22400
\u89e3\u5f97M=33600\uff08a-c\uff09/b
氨气的物质的量为700/22.4=31.25mol
氨气的质量为17×31.25=531.25g
氨气的质量分数为531.25/(1000+531.25)=34.7%
(2)1L水中溶有标况下VL气体摩尔质量为Mg/mol生成密度为bg/ml的溶液,其公式为1000bV/(22400+MV),由此可以得出结果。
1、700/22.4*17÷(700*17+1000)*100%=531.25/1531.25=34.7
700÷22.4 NH3的物质的量 m=n*M
2 a÷22.4 HCL体积 HCl气体质量 36.5a/22.4 溶液质量 36.5a/22.4+1000 体积为质量÷密度
得到的体积单位是ml 因为c=n/V V要求得是L 所以会有1000
(1)设有1L水,则氨气为700L。氨气的质量=700*22.4*17=531.25克
则溶质的质量分数=531.25/(531.25+1000)=34.7%
(2)n(HCl)=a/22.4 mol
溶液体积=溶液质量/溶液密度=(a/22.4 *36.5+1000)/1000b
所以盐酸的量浓度=n/V=a/22.4 /(a/22.4 *36.5+1000)/1000b=1000ab/(22400+36.5a)
绛旓細(2)1L姘翠腑婧舵湁鏍囧喌涓媀L姘斾綋鎽╁皵璐ㄩ噺涓篗g/mol鐢熸垚瀵嗗害涓篵g/ml鐨勬憾娑诧紝鍏跺叕寮忎负1000bV/(22400+MV),鐢辨鍙互寰楀嚭缁撴灉銆
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绛旓細涓銆佹懇灏旓細 銆1锛氭懇灏旀槸鍋氫负璁¢噺 鍘熷瓙銆佸垎瀛 绛夊井瑙傜矑瀛愮殑鐗╄川鐨勯噺鐨勫崟浣嶃2锛氱墿璐ㄧ殑閲忕殑绗﹀彿涓 mol 銆傛懇灏旂殑绗﹀彿涓 n 銆3锛1 mol浠讳綍绮掑瓙鐨勭矑瀛愭暟鍙 闃夸紡鍔犲痉缃楀父鏁 銆傜鍙蜂负 NA 銆傞氬父浣跨敤鐨勮繎浼煎间负 6.02*10^23 銆4锛氱墿璐ㄧ殑閲...
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绛旓細31銆佹爣鍐典笅瀵嗗害涓1.25g/L,m=蟻v.鍒1mol璇ユ皵鎬佺儍鐨勮川閲忎负1.25g/L*22.4L/mol=28g/mol銆傚垯璇ョ儍鎽╁皵璐ㄩ噺涓28g/mol銆1mol鐑冧腑纰崇殑璐ㄩ噺浣28*85.7%=23.996鈮24锛屾阿鐨勮川閲28*14.3=4.004鈮4銆傚垎瀛愬紡涓篊2H4