初三化学两道计算题~谢谢谢谢!!!!
\u4e24\u9053\u521d\u4e09\u5316\u5b66\u8ba1\u7b97\u98981\u3001Fe3O4\u2014\u20144CO2
232 4\u00d744
x 5.5g
x=232 / 32
\u8d28\u91cf\u5206\u6570= (x / 10) \u00d7100%=72.5%
Fe3O4\u2014\u20144Fe
232 4\u00d756
x 12.8g
x=232 \u00d712.8 / 224
\u8d28\u91cf\u5206\u6570= (x / 16) \u00d7100%=82.9%
\u521d\u4e2d\u7684\u554a
1.150T=150000000g \u8bbe\u6c27\u6c14\u8d28\u91cf\u4e3aX
CH4(m)=150000000 X 0.8=120000000g
CH4+2O2=CO2\uff08\u4e0a\u7bad\u5934\uff09+2H20
16 64
120000000g x
16\120000000=64\x
x=480000000g
V(o2)=m/\u5bc6\u5ea6=34285714.26L
2\uff0cCO2+CA(OH)2=CaCO3+H2O
CaCO3(m)=37.5g
\u7b97\u51faCO2\u8d28\u91cf
\u4ece\u800c\u7b97\u51faCO\u7684\u8d28\u91cf
Fe(x)O(y)--------Y/XCO\u4e2d\uff0c\u6c27\u5316\u7269Fe\u4e3a 2Y/X\u4ef7
56x+16y Y \u4e5828
2\u300220g CO\u8d28\u91cf
\u5f97\u5230Y/X.\u5c31\u5f97\u5230FE\u7684\u5316\u5408\u4ef7\uff0c\u5c31\u786e\u5b9a\u5316\u5b66\u5f0f\u4e86
\u521d\u4e2d\u5f88\u8bb0\u4e0d\u5f97\u4e86\uff0c\u601d\u8def\u5e94\u8be5\u5c31\u662f\u8fd9\u4e2a\uff0c\u6709\u4ec0\u4e48\u5730\u65b9\u4e0d\u5bf9\u539f\u8c05\u4e0b\u54c8
2HCl+Na2CO3==2NaCl+H2O+CO2↑
106 117 44
m1 m2 8.8g
106/m1 = 117/m2 = 44/8.8
得m1 = 21.2g, m2 = 23.4g
所以原固体中NaCl质量 = 202.9-32.9-21.2 = 148.8g
加上生成的23.4g NaCl
所以反应后溶液中NaCl质量 = 148.8 + 23.4 = 172.2g
溶液质量 = 521.0 - 32.9 = 488.1g
所得溶液中溶质的质量分数 = 172.2/488.1 = 35.3%
根治质量守恒，生成的BaCO3沉淀质量 = 100+153.7-234 = 19.7g
Na2CO3+BaCl2==BaCO3↓+2NaCl
197 117
19.7g m
197/19.7 = 117/m
得m = 11.7g
所以过滤所得溶液中溶质的质量分数 = 11.7/234 = 5%
1、解：（1）由题意分析知，生成二氧化碳的质量为：202.9g+326.9g-521.0g=8.8g
（2）设碳酸钠，反应生成的氯化钠质量分别为x、y
Na2CO3+2HCl=2NaCl+H2O+CO2
106 11 7 44
X Y 8.8g
算式略
x=21.2g y=23.4g
（3）氯化钠的质量分数为：（32.9g-21.2g+23.4g）/521.0g*100%=6.7%
答：略
2、 解：（1）由题意分析知，生成碳酸钡的质量为：100g+153.7g-234g=19.7g
（2）设反应生成氯化钠质量分别为y
Na2CO3+BaCl2=2NaCl+BaCO3
11 7 197
Y 19.7g
算式略
y=11.7g
（3）氯化钠的质量分数为：11.7g/234g*100%=5%
答：略
绛旓細1:鍔犲叆鐩愰吀鐢变簬2HCl+Na2CO3=H2O+NaCl+CO2鍙嶅簲锛屼骇鐢熺殑鐗╄川浠呮湁CO2閫稿嚭鐑ф澂锛屽洜姝璁＄畻鐨勬瘡鍔犲叆73鍏婬Cl锛岀儳鏉川閲忓鍔犺川閲忔槸29鍏嬨傝鍔犲叆10%鐨勭洂閰竂鍏嬶紝鍒欏疄闄呯儳鏉鍔犺川閲忎负锛0.1X/73*29+0.9X=686X/730 寰鐑ф澂涓姞鍏ユ隘鍖栭挋锛岀敓鎴愮殑鏄⒊閰搁挋娌夋穩锛屽洜姝ゅ姞鍏1鍏嬫隘鍖栭挋锛岀儳鏉閲1鍏 鐢变簬澶╁钩鏈缁堝钩琛...
绛旓細y=8.8g 2)鐭崇伆鐭虫牱鍝佷腑纰抽吀閽欑殑璐ㄩ噺鍒嗘暟涓猴細20g/25g*100%=80%.杩欐槸鍒╃敤鍖栧鏂圭▼寮忕畝鍗璁＄畻銆傚叧閿槸鎶撲綇鍏崇郴銆
绛旓細KClO3鐨勫垎瑙ｇ巼涓 1.12a/2.77a*100 =40.5
绛旓細1 19%鏉傝川 灏辨湁81%鐨凢e2O3 Fe2O3鍙堟湁56*2/(56*2+16*3)*100%=70% 鐨凢e 鎵浠e鏈8*81%*70%=4.536t 鍙堣姹傞搧涓96 璁剧敓閾乆t 鏈4.536/X=96 X=4.725t 2 鐩愰吀鍜岄粍閾滃悎閲戠殑鎬诲拰鐭ラ亾 閭ｄ箞鍙嶅簲鍚庡噺灏戠殑璐ㄩ噺灏辨槸鏀惧嚭鐨勬阿姘 鎵浠ユ阿姘旇川閲:16.25+85.00-101.11=0.14g 鍙湁閿屼笌鐩...
绛旓細=14/79 m锛堢⒊閾碉紙NH4HCO3锛夛級=0.28/锛14/79锛=1.58g 绛旓細闇璐拱1.58g鐨勭⒊閾碉紙NH4HCO3锛夈2. 璇ュ弽搴旂殑鍖栧鏂圭▼寮忔槸 2KClO3===2KCL+3O2鈫 锛=== 涓婇潰鍐欐潯浠垛斺斿姞鐑拰浜屾哀鍖栭敯锛夊弬鍔犲弽搴旂殑姘吀閽惧拰姘ф皵鐨勮川閲忔瘮涓 245锛96 24.5/245=x/96 x=9.6g 绛旓細鍙埗鍙栨哀姘9.6g銆
绛旓細CaCO3+2HCl==CaCl2+CO2(姘斾綋)+H2O 100 73 111 44 x y z 2.86g 100/44=x/2.86g x=6.5g 73/44=y/2.86g y=4.745g 111/44=z/2.86g z=7.215g 1銆丆aCO3%=6.5g/10g*100%=65 2銆丠Cl%=4.745g/90g*100%=5.3 3銆佸畬鍏ㄥ弽搴斿悗鎵寰楁憾娑茬殑婧惰川璐...
绛旓細2HCl+Na2CO3==2NaCl+H2O+CO2鈫 106 117 44 m1 m2 8.8g 106/m1 = 117/m2 = 44/8.8 寰梞1 = 21.2g, m2 = 23.4g 鎵浠ュ師鍥轰綋涓璑aCl璐ㄩ噺 = 202.9-32.9-21.2 = 148.8g 鍔犱笂鐢熸垚鐨23.4g NaCl 鎵浠ュ弽搴斿悗婧舵恫涓璑aCl璐ㄩ噺 = 148.8 + 23.4 = 172.2g 婧舵恫璐ㄩ噺 = 521...
绛旓細姘寸殑璐ㄩ噺=5.4g 瑙ｏ細璁剧敳鐑疯川閲忎负x 16 36 x 5.4g 16 36 ___ =___x 5.4g x= 2.4g 鈭寸敳鐑疯川閲忎负2.4g
绛旓細璁剧敓鎴愮⒊閰搁挋x鍏嬶紝鐢熸垚姘寲閽緔鍏嬶紝鏍规嵁鍙嶆槧鏈塁aCl2+K2CO3=CaCO3鈫+2KCl 138 100 149 13.8 x y 138 锛100=13.8 锛歺 138 锛149 = 13.8 锛 y 瑙ｅ緱x=10鍏嬶紝y=14.9鍏 鍗崇敓鎴愮⒊閰搁挋娌夋穩璐ㄩ噺涓猴細10鍏嬶紝鏈鍚庢墍寰楁憾娑蹭腑婧跺墏锛堟按锛夌殑鎬昏川閲忎负锛13.8+26.2=40鍏嬶紝鏍规嵁T...
绛旓細Y X 1.6 X= 14KG Y= 40KG c%= 40/100= 40 (1)鍑忓皬鐨勮川閲忎负姘у寲閾滀腑姘у厓绱犵殑璐ㄩ噺銆俋*16/80= 1.6 X= 8g (2)CO+CuO=CO2+Cu 80 44 8 Y Y=4.4g CO2+Ca(OH)2=CaCO3+H2O 44 100 4.4 Z Z=10g ...
