一道高一化学题，帮帮忙，谢谢！ 一道高中化学题，帮帮忙，谢谢~~~
\u4e00\u9053\u9ad8\u4e00\u5316\u5b66\u9898\uff0c\u5e2e\u5e2e\u5fd9\uff0c\u8c22\u8c22O(\u2229_\u2229)O\uff01\uff081\uff09\u6839\u636e\u94dd\u571f\u77ff\u4e2dFE2O3\u5168\u90e8\u8f6c\u5165\u8d64\u6ce5
m\uff08\u94dd\u571f\u77ff\uff09*16%=m\uff08\u8d64\u6ce5\uff09*48%
\u89e3\u5f97m\uff08\u8d64\u6ce5\uff09=1/3m\uff08\u94dd\u571f\u77ff\uff09
m\uff08\u94dd\u571f\u77ff\uff09=1t\u65f6\uff0c\u89e3\u5f97m\uff08\u8d64\u6ce5\uff09=0.33t
\u751f\u4ea7\u4e2d\u6bcf\u6d88\u80171t\u94dd\u571f\u77ff\uff0c\u5c06\u4ea7\u751f0.33\u5428\u8d64\u6ce5\u3002
\uff082\uff09m(Al)=m(Al2O3)*[2M(Al)/M(Al2O3)]
=[m\uff08\u94dd\u571f\u77ff\uff09*55%-m\uff08\u8d64\u6ce5\uff09*15%]*[2M(Al)/M(Al2O3)]
=[m\uff08\u94dd\u571f\u77ff\uff09*55%-1/3m\uff08\u94dd\u571f\u77ff\uff09*15%]*[2M(Al)/M(Al2O3)]
=[m\uff08\u94dd\u571f\u77ff\uff09*50%]*[2M(Al)/M(Al2O3)]
=m\uff08\u94dd\u571f\u77ff\uff09*M(Al)/M(Al2O3)
m\uff08\u94dd\u571f\u77ff\uff09=m(Al)*M(Al2O3)/M(Al)
=1t*(27*2+16*3)/27
=3.78t
m(NaOH)=n(NaOH)*M(NaOH)
=n(Na)*M(NaOH)
=2n(Na2O)*M(NaOH)
=2m(Na2O)/M(Na2O)*M(NaOH)
=2m(\u8d64\u6ce5)*8%/M(Na2O)*M(NaOH)
=2*1/3m\uff08\u94dd\u571f\u77ff\uff09*8%/M(Na2O)*M(NaOH)
=2/3*3.78t*8%/(23*2+16)*(23+16+1)=0.13t
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(1) 已知铝土矿中FE2O3全部转入赤泥, 每消耗1吨铝土矿, 产生1*16% = 0.16吨Fe2O3赤泥中的Fe2O3含量为48%, 赤泥质量: 0.16/48% = 0.33吨
(2) 赤泥中的Al2O3含量为15%,0.33吨赤泥中含Al2O3质量: 0.33*15% = 0.05吨
铝土矿的Al2O3含量为55%, 1吨铝土矿中含Al2O3质量为: 1*55% = 0.55吨
用于电解制铝的Al2O3质量为: 0.55-0.05 = 0.50吨
2Al2O3===电解====4Al+3O2
2*102 4*27
0.50 x
x = 4*27*0.50/(2*102) = 0.26吨
1吨铝土矿可得0.26吨铝,设每炼出1吨铝理论上需铝土矿的质量为y吨:
1 : 0.26 = y : 1
y = 1*1/0.26 = 3.85吨
每消耗1吨铝土矿产生0.33吨赤泥, 其中含Na2O: 0.33*8% = 0.026吨
Na2O ----- 2NaOH
62 2*40
0.026 z
z = 2*40*0.026/62 = 0.034吨
设每炼出0.26吨铝, 应补充0.034吨NaOH; 设每炼出1吨铝应补充w吨NaOH:
0.26 :0.034 = 1 : w
w = 0.034*1/0.26 = 0.13
(1) 已知铝土矿中FE2O3全部转入赤泥, 每消耗1吨铝土矿, 产生1*16% = 0.16吨Fe2O3
赤泥中的Fe2O3含量为48%, 赤泥质量: 0.16/48% = 0.33吨
(2) 赤泥中的Al2O3含量为15%,0.33吨赤泥中含Al2O3质量: 0.33*15% = 0.05吨
铝土矿的Al2O3含量为55%, 1吨铝土矿中含Al2O3质量为: 1*55% = 0.55吨
用于电解制铝的Al2O3质量为: 0.55-0.05 = 0.50吨
2Al2O3===电解====4Al+3O2
2*102 4*27
0.50 x
x = 4*27*0.50/(2*102) = 0.26吨
1吨铝土矿可得0.26吨铝,设每炼出1吨铝理论上需铝土矿的质量为y吨:
1 : 0.26 = y : 1
y = 1*1/0.26 = 3.85吨
每消耗1吨铝土矿产生0.33吨赤泥, 其中含Na2O: 0.33*8% = 0.026吨
Na2O ----- 2NaOH
62 2*40
0.026 z
z = 2*40*0.026/62 = 0.034吨
设每炼出0.26吨铝, 应补充0.034吨NaOH; 设每炼出1吨铝应补充w吨NaOH:
0.26 :0.034 = 1 : w
w = 0.034*1/0.26 = 0.13
这个回答比较详细了，就是如果题型有变化的时候，要知道先分辨出那个量少，以那个做基准的分析。
绛旓細1)姘寲閾电殑鐗╄川鐨勯噺=107/锛14+4+35.5锛=2mol 2NH4Cl+Ca(OH)2=CaCl2+2NH3+2H2O 2 2 2 x x=2*2/2=2 浜х敓鐨凬H3鐨勭墿璐ㄧ殑閲=2mol 鏍囧喌涓嬭繖浜涙皵浣撶殑浣撶Н=2*22.4=44.8鍗 2锛夋皑姘寸殑鐗╄川鐨勯噺娴撳害=2/0.1=20mol/L
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绛旓細浣15.55鍏嬪潎鏄隘鍖栭挔 璁 婧村寲閽犵殑璐ㄩ噺涓篗锛屽垯 鍘熸潵姘寲閽犵殑璐ㄩ噺涓20-M 鏍规嵁閽犲厓绱犲畧鎭掞紙鍓嶅悗鍗充笉澧炰篃閮ㄥ噺锛夊彲鍒楁柟绋嬪紡 23/103 *M + 锛20-M锛 *23/58.5 ==15.55 *23/58.5 瑙ｆ柟绋嬪緱 M=10.3鍏 閭ｄ箞婧寸殑璐ㄩ噺=10.3*80/103=8鍏 鎵浠 璐ㄩ噺鍒嗘暟=8/20=40 搴旇娌￠敊 ...
