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# 综合计算 - PowerPoint PPT Presentation

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x = 175g

n (NaNO3) = (175g) (22%) / 85g · mol-1 = 0. 45mol

C (NaNO3) = 0.45mol / 0.15L = 3 mol/L

CuSO4 + Ba(NO3)2 = BaSO4 + Cu(NO3)2

0. 02mol

0. 02mol

0. 02mol

HCl + AgNO3 = AgCl + HNO3

0. 15mol

0. 15mol

0. 15mol

NaOH + HNO3 = NaNO3 + H2O

0. 15mol

0. 15mol

2NaOH + Cu(NO3)2 = 2NaNO3 + Cu(OH)2

0. 04mol

0. 02mol

0.15mol + 0.04mol

= 0.19L = 190mL

x =

1mol /L

Cu + 2AgNO3 = 2Ag + Cu(NO3)2

△m

64g

2mol

2×108g

(2×108 )g - 64g = 152g

x

10.152g - 10g = 0.152g

2mol : x = 152g : 0.152g x = 0.002mol

0.042mol – 0.002mol = 0.04mol

NaCl + AgNO3 = NaNO3 + AgCl

KCl + AgNO3 = KNO3 + AgCl

NaCl + AgNO3 = NaNO3 + AgCl

x

x

KCl + AgNO3 = KNO3 + AgCl

y

y

x + y = 0.04mol

x = 0.03 mol

(58.5g/mol)x + (74.5g/mol)y = 2.5g

y = 0.01 mol

NaCl和KCl 的质量为 2. 5g

NaCl和KCl 混合物的平均摩尔质量为 2. 5g/0.04mol = 62.5g/mol

NaCl 58.5

12

3

62.5

KCl 74.5

4

1

n(NaCl) =(0.04mol)(3/4) = 0. 03mol

n(KCl) = (0.04mol)(1/4) = 0. 01mol

n(MnO2) = 8.70g / 87g · mol-1 = 0.1mol

n(HCl) = (125g)(36. 5%)/36. 5g · mol-1 = 1.25 mol

MnO2 + 4HCl = MnCl2 + 2H2O + Cl2

1mol

4mol

0.1mol

1.25mol

∵ 1×1.25 > 4×0.1

∴ 盐酸过量

n(Cl2) = 0.1mol

HCl + AgNO3 = HNO3 +AgCl

MnCl2 + 2AgNO3 = Mn(NO3)2 + 2AgCl

200mL溶液中 n(Cl-) = (1.05 mol)(1/5) = 0.21mol

m(AgCl) = ( 0.21mol) (143.5g/mol) = 30.1g

1. 写出有关变化的化学方程式

2. 找出已知条件

3. 分析列式计算。

4. 常用的计算方法：

2000年 11月16日