CE CHEM--CH.18

1.(i)
3NaOH + H3PO4 → Na3PO4 + 3H2O

Mole ratio NaOH : H3PO4 = 3 : 1
No. of moles of H3PO4 used = MV = 1.5 x (50/1000) = 0.075 mol
No. of moles of NaOH used = 0.075 x 3 = 0.225 mol
Vol. of NaOH = mol/M = 0.225/2 = 0.1125 dm3

Mole ratio H3PO4 : Na3PO4 = 1 : 1
No. of moles of H3PO4 used = 0.075 mol
No. of moles of Na3PO4 formed = 0.075 mol
Final volume of the solution = (50/1000) + 0.1125 = 0.1625 dm3
Molarity of H3PO4 = mol/V = 0.075/0.1625 = 0.4615 M

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2.(i)
2HCl + Pb(NO3)2 → PbCl2↓ + 2HNO3
Mole ratio Pb(NO3)2 : PbCl2 = 1 : 1
No. of moles of Pb(NO3)2 used = MV = 0.5 x (4/1000) = 0.002 mol
No. of moles of PbCl2 formed = 0.002 mol

Molar mass of PbCl2 = 207.2 + 35.5x2 = 278.2 g mol-1
Mass of PbCl2 formed = mol x (molar mass) = 0.002 x 278.2 = 0.5564 g

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2(ii)
1 mole of PbCl2 contains 6.02 x 1023 formula units.
No. of moles of PbCl2 formed = 0.002 mol
No. of fomula units of PbCl2 formed = 0.002 x (6.02 x 1023) = 1.204 x 1021

2008-04-26 22:51:34 補充：
1.(i)
The reaction is neutralization. It is impossible to form H2 in acid-alkali neutralization.

2(ii)
The white precipitate formed is PbCl2, but NOT Pb(NO3)2.

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