我问一道高一的化学计算题（我是即将高一的新生，请帮帮忙！）
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n\uff1dN/NA
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2 H2 + O2=2 H2O
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m克CuSO4·5H2O中含有CuSO4质量 = m/250 * 160 = 16m/25
m克CuSO4·5H2O中含有水质量 = m/250 * 90 = 9m/25
加上蒸发掉的80克水，溶液一共损失了水的质量 = 9m/25 + 80
9m/25 + 80 克水能溶解的CuSO4的质量最大值 = (9m/25 + 80)/100 * 20 = 9m/125 + 16
显然有：损失的水所能溶解的CuSO4的最大量 = 析出的CuSO4.5H2O中CuSO4的质量
所以有：16m/25 = 9m/125 + 16
m = 2000/71
所以析出CuSO4·5H2O质量为 2000/71 克
结晶水合物的析晶计算难度大，是由于带有结晶水晶体的析出，会导致溶剂水量的减少，从而使结晶水合物继续从饱和溶液中析出，这样依次重复，最终使晶体的总量趋向于定值。由此可见，结晶水合物的析出过程实质上是无数次结晶的总结果。作为一个数学问题，这类题目可以应用无穷递缩等比数列求和知识解决，但初中学生尚未学过，故对于学生来说是陌生的。若仔细分析题意，可抓住析晶后的溶液仍为饱和溶液求解。
原饱和溶液中溶质CuSO4的质量=300g×20g/120g=50g。水的质量=300g－50g=250g。
设析出CuSO4·5H2O质量为x.
CuSO4·5H2O ~ CuSO4 ~ 5H2O
250 160 90
x 160/250x 90/250x
20g/100g=（50g－160/250x）/（250g－80g－90/250x）
解得x=28.2克
希望对你有帮助 ↖(^ω^)↗
绛旓細m鍏婥uSO4路5H2O涓惈鏈塁uSO4璐ㄩ噺 = m/250 * 160 = 16m/25 m鍏婥uSO4路5H2O涓惈鏈夋按璐ㄩ噺 = m/250 * 90 = 9m/25 鍔犱笂钂稿彂鎺夌殑80鍏嬫按锛屾憾娑蹭竴鍏辨崯澶变簡姘寸殑璐ㄩ噺 = 9m/25 + 80 9m/25 + 80 鍏嬫按鑳芥憾瑙ｇ殑CuSO4鐨勮川閲忔渶澶у = (9m/25 + 80)/100 * 20 = 9m/125 + 16 鏄剧劧鏈夛細...
绛旓細8.7g m 7.1 m m=7.1g m=25.4 n(I2)=25.4梅254=0.1mol I2锛2Na2S2O3锛2NaI锛婲a2S4O6 1 2 0.1 n n=0.2mol V(Na2S2O3)=0.2梅1=
绛旓細(2)c(Na+)=n鎬(Na+)/V=0.2mol/0.1L=2mol/L 绛旓細(1)鍘熸贩鍚堢墿涓璑aCl鐨勮川閲忓垎鏁颁负53.92 (2)鍒舵垚鐨100ml娣峰悎娑蹭腑锛孨a绂诲瓙鐨勭墿璐ㄧ殑閲忔祿搴︿负2mol/L.
绛旓細2)鐢遍鎰忥紝璁惧師娣峰悎鐗╀腑鏈夌⒊閰搁挔Xmol锛岀⒊閰告阿閽燳mol 鍒楁柟绋嬬粍锛106X+84Y=27.4鍏 (X+Y\2)*106=21.2鍏 寰梄=0.1 Y=0.2 鍒欑⒊閰搁挔鏈10.6鍏嬬⒊閰告阿閽16.8鍏 褰撳姞鐑埌璐ㄩ噺涓嶅湪鍙樺寲鏃跺嵆鏈夌⒊閰搁挔21.2鍏嬶紝涓0.2mol 鏍规嵁鍙嶅簲寰楃洂閰告懇灏旀暟涓0.4mol 鎵浠ョ洂閰哥墿璐ㄧ殑閲忔祿搴︿负0.4\0.2...
绛旓細锛1锛2NaOH + H2SO4===2H2O + Na2SO4 NaCO3 + H2SO4===H2O + CO2鈫+ Na2SO4 鍥犱负瀹屽叏鍙嶅簲,鎵浠ュ彲浠ユ牴鎹～閰告牴绂诲瓙鐨勬懇灏旇川閲忔潵璁＄畻纭吀閽犵殑璐ㄩ噺锛岀～閰哥殑鎽╁皵璐ㄩ噺涓98*20%/98=0.2mol 鎵浠ョ～閰搁挔鐨勮川閲忎负0.2*142=28,4g 锛2锛夋牴鎹川閲忓畧鎭掑師鐞嗭紝鐢熸垚鐨勬矇娣鐨勮川閲忓氨鏄弽搴斿墠鐨勬昏川閲忓噺...
绛旓細1銆佺閰搁摱涓庣洂閰稿強鍙憾鎬х洂閰哥洂婧舵恫:Ag++Cl-=AgCl鈫2銆侀挔涓庢按鍙嶅簲:2Na+2H2O=2Na++2OH鈥+H2鈫3銆侀挔涓庣～閰搁摐婧舵恫:2Na+2H2O+Cu2+=2Na++Cu(OH)2鈫+H2鈫4銆佽繃姘у寲閽犱笌姘村弽搴:2Na2O+2H2O=4Na++4OH鈥+O2鈫5銆佺⒊閰告阿鐩愭憾娑蹭笌寮洪吀婧舵恫娣峰悎:HCO3-+H+=CO2鈫+H2O6銆佺⒊閰告阿鐩愭憾娑蹭笌閱嬮吀婧舵恫娣峰悎:HCO3-+...
绛旓細NaHCO3锛+2路n锛圢a2CO3锛=0.15 mol 鎵浠锛圚Cl锛=36.5路0.15=5.475 g 璐ㄩ噺鍒嗘暟=5.475/50=0.1095 3锛変骇鐢熺殑CO2鐨勮川閲=10/100路44=4.4 g m锛圢aCl锛=58.5路n锛圢aCl锛=58.5路0.15=8.775 g 婧舵恫鎬昏川閲=9.5+50-4.4=55.1 g 鎵浠ユ墍寰楁憾娑茬殑璐ㄩ噺鍒嗘暟=8.775/55.1=0.159 ...
绛旓細绛旀锛72.3 25锛庡垎鏋愶細CO2涓嶯a2O2鍙嶅簲锛屼互鍙奌2O涓嶯a2O2鍙嶅簲鍧囦娇鍥轰綋鐗╄川璐ㄩ噺澧炲姞锛岃繖涓庨缁欏浐浣撹川閲忎粠37.16 g鍙樹负40 g锛屽鍔2.84 g鐩稿叧锛屽彲杩愮敤鍖栧鏂圭▼寮忚繘琛屸滆川閲忓樊閲忔硶鈥濆垪寮璁＄畻鍑轰袱绉嶆皵浣撶殑鐗╄川鐨勯噺銆傚彟澶栵紝4.76 g+37.16 g锛40 g=1.92 g锛1.92 g鍗虫槸鐢熸垚鐨勬哀姘旂殑璐ㄩ噺锛屽悓鏍峰彲浠...
绛旓細2Na+2H2O=2NaOH+H2鈫 46 80 2 x y 0.2g 46/x=80/y=2/0.2 x=4.6g y=8g Na2O鐨勮川閲忔槸5.22-4.6=0.62g Na2O+H2O=2NaOH 62 80 0.62g z 62/0.62=80/z z=0.8g 鍙嶅簲鐢熸垚NaOH婧惰川鐨勮川閲忔槸8+0.8=8.8g 鍙嶅簲鍚庢憾娑茬殑璐ㄩ噺鏄5.22+94.98-0.2=100g NaOH...
绛旓細V锛圕l2锛=14.2/71*22.4=4.48L H2 + Cl2 = 2HCl 22.4L 22.4L 44.8L 11.2L 4.48L V锛圚Cl锛夋牴鎹鎰忥紝姘㈡皵涓庢隘姘斿簲绛変綋绉弽搴旓紝鎵浠ユ阿姘旀湁鍓╀綑锛屽墿浣6.72L 鐢熸垚姘寲姘㈢殑浣撶Н涓8.96L c锛圚Cl锛=(8.96/22.4)mol/0.25L=1.6mol/L ...
