Some urgent questions

1.
Denote the metal ion as Mn+.
Since chloride ion is Cl-, the formula of the metal chloride is MCln.

No. of moles of Mn+ = MV = 0.1 x (50/1000) = 0.005 mol
No. of moles of Cl- = MV = 0.1 x (100/100) = 0.01 mol
Mole ratio Mn+ : Cl- = 0.005 : 0.01 = 1 : 2
From the formula MCln, mole ratio Mn+ : Cl- = 1 : n
Hence, n = 2

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2.
Molar mass of NaOH = 23 + 16 + 1 = 40 g mol-1
Molar mass of Na­2CO4 = 23x2 + 12 + 16x3 = 106 g mol-1

Let the mass of NaOH = y g
Then the mass of Na2CO3 = (1.5 - y) g

Both NaOH and Na2CO3­ react with HCl :
NaOH + HCl → NaCl + H2O …. (1)
Na2CO3 + 2HCl → 2NaCl + H2O + CO2 …. (2)

No. of moles of NaOH = mass/(molar mass) = (y/40) mol
Refer to (1).
No. of moles of HCl reacted with NaOH = (y/40) mol
No. of moles of Na2CO3 = mass/(molar mass) = [(1.5-y)/106] mol
Refer to (2).
No. of moles of HCl reacted with Na2CO3 = 2[(1.5-y)/106] mol
Refer to both (1) and (2).
Total no. of moles of HCl = {(y/40) + 2[(1.5-y)/106]} mol
But total no. of moles of HCl = MV = 1 x (30/1000) = 0.03 mol

Therefore, (y/40) + 2[(1.5-y)/106] = 0.03
Mass of NaOH = y = 0.277 g
% by mass of NaOH = (0.277/1.5) x 100% = 18.5%
% by mass of Na2CO3 = (100 - 18.5)% = 81.5%

2008-03-01 13:13:57 補充：
In Question 1:
Since n = 2, the charge on the metallic ion = +2

在 Lilian 的答案（回答 001）中，第 2 題的做法是認為 Na2CO3 不與 HCl 反應。但其實 Na2CO3 與 HCl 是有反應的，所以 80% 和 20% 的答案是錯的。

1. the no. of mole of silver ions in silver nitrate solution = molarity x volume
volume = 10cm^3 = 0.01dm^3
no. of mole of silver ions = 0.1 x 0.01 = 0.001
= no. of mole of chloride ions

no. of mole of metallic chloride = 0.1 x 0.005 = 0.0005
charges on the metallic ion = 0.001/ 0.0005 = 2

2. molar mass of NaOH = 23 + 16 + 1 = 40
molar mass of Na2CO3 = 23 x2 + 12 + 16 x 3 = 106
no. of mole of HCl used = 1M x 0.03dm^3 = 0.03 = no. of mole of NaOH
mass of NaOH = 40 x 0.03 = 1.2g
% composition by mass of sodium hydroxide = 1.2/ 1.5 = 80%
% composition by mass of sodium carbonate = 1- 80% =20%