Chem...involving titrations

1)
(1) Mg + H2SO4 →MgSO4 + H2
(2) MgO + H2SO4 → MgSO4 + H2O
(3) 2NaOH + H2SO4 → Na2SO4 + 3H2O

In (1):
No. of moles of H2 used = 0.15 mol
No. of moles of Mg used = 0.15 mol
No. of moles of H2SO4 reacted with Mg = 0.15 mol

In (3):
No. of moles of NaOH used = 0.2 mol
No. of moles of H2SO4 reacted with NaOH = 0.2 x (1/2) = 0.1 mol

In (2):
Total no. of H2SO4 used = 3 x (100/1000) = 0.3 mol
No. of moles of H2SO4 reacted with MgO = 0.3 - 0.15 - 0.1 = 0.05 mol
No. of moles of MgO = 0.05 mol

In the sample:
Mass of Mg = 0.15 x 24.3 = 3.645 g
Mass of MgO = 0.05 x (24.3 + 16) = 2.015 g
% purity of Mg = [3.645/(3.645 + 2.015)] x 100% = 64.4%


2)
Consider the reaction between the resultant solution and NaOH:
HCl + NaOH → NaCl+ H2O
No. of moles of NaOH used = 1 x (30/1000) = 0.03 mol
No. of moles of excess HCl = 0.03 mol

XCO3 + 2HCl → XCl2 + H2O + CO2
Total no. of moles of HCl = 1 x (50/1000) = 0.05 mol
No. of moles of HCl reacted with XCO3 = 0.05 - 0.03 = 0.02 mol
No. of moles of XCO3 = 0.02 x (1/2) = 0.01 mol
Molar mass of XCO3 = 2.5/0.01 = 250 g/mol
Relative atomic mass of X = 250 - 12 - 16x3 = 190


3)
Consider the neutralization:
H3X + 3NaOH → Na3X + 3H2O
No. of moles of NaOH used = 0.3 x (28.6/1000) = 0.00858 mol
No. of moles of H3X in 25 cm³ solution = 0.0858 x (1/3) = 0.00286 mol

No. of moles of H3X in 250 cm³ solution = 0.00286 x (250/25) = 0.0286 mol
Molar mass of H3X = 2.8/0.0286 = 97.9 g/mol
Relative molecular mass of H3X = 97.9

2010-06-15 17:18:44 補充：
Question 1 has many steps because there are 3 reactions.

In the Mg sample, the impurity is MgO.
Mass of Mg = 3.645 g
Mass of MgO = 2.015 g.
Total mass of the sample = (3.645 + 2.015) g

% purity of Mg
= [(Mass of Mg)/(Total mass of sample)] x 100%
= [3.645/(3.645 + 2.015)] x 100%
= ........