chem f.4

Consider the neutralization of 25 cm3 Na2CO3 solution against HCl:
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
Mole ratio Na2CO3 : HCl = 1 : 2
No. of moles of HCl used = 0.098 x (18.2/1000) = 0.0017836 mol
No. of moles of Na2CO3 used = 0.0017836 x (1/2) = 0.0008918 mol

Consider the preparation of 250 cm3 of Na2CO3 solution:
No. of moles of Na2CO3 in 25 cm3 solution = 0.0008918 mol
No. of moles of Na2CO3 in 250 cm3 solution = 0.0008918 x (250/25) = 0.008918 mol
No. of moles of Na2CO3•nH2O to make the solution = 0.008918 mol
Mass of Na2CO3•nH2O to make the solution = 2.549 g
Molar mass of Na2CO3•nH2O = 2.549/0.008918 = 286 g/mol

Consider the molar mass of Na2CO3•nH2O:
23x2 + 12 + 16x3 + n(1x2 + 16) = 286
106 + 18n = 286
n = 10
=

Firstly, write down the equation.

Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)

1/2 X no.of moles of HCl = no.of moles of Na2CO3

no. of moles of Na2CO3 = 18.2/1000 X 0.098 / 2
= 8.918 X 10^-4 mol

∵ As Na2CO3.nH2O was dissolved in distilled water and made up to 250.0 cm3 and then pipette 25cm3 of the solution required 18.2 cm3 of 0.098 M of HCl for complete reaction

∴ no. of moles of Na2CO3 in 25cm3 solution = 2.549(10) / 18n X 106

Thus,

no.of moles of HCl added = 2 X no.of moles of Na2CO3
2.549(10) / 18n X 106 = 8.918 X 10^-4
n = 14.98
n ~ 15

2009-05-13 22:39:25 補充：
唔好意思 計錯左數 老爺子先對