兩條有難度的化學摩爾算題

H2C2O4 + 2NaOH --> Na2C2O4 + 2H2O
no of mole of NaOH = 0.1X(14.75/1000) = 1.475X10^-3
no of mole of H2C2O4 = 1.475X10^-3 / 2 = 7.375X10^-4

5(C2O4)2- + 2MnO4- + 16H+ --> 10CO2 + 2Mn2+ + 8H2O
no of mole of MnO4- = 0.0205X(30.5/1000) = 6.02525X10^-4
no. of mole of (C2O4)2- = 6.02525X10^-4 X ( 5/2 ) = 1.563125X10^-3

Conc. of H2C2O4 = 7.375X10^-4 / (25/1000) = 0.0285 M
Conc. of 草酸鈉 (Na2C2O4) = (1.563125X10^-3 - 7.375X10^-4) / (25/1000) = 0.0330 M

2008-09-27 22:24:45 補充：
(S2O3)2- + I2 -> (S4O6)2- + 2I
2 Cu^2+ + 4 I^- ---> 2 Cul + l2
0.499克硫酸銅結晶(CuSO4.5H2O) = 0.499 / (63.5+32+16x4+518) = 0.002 mole

2008-09-27 22:32:55 補充：
KI in excess, no S2O3^2- is needed.

2(S2O3)2- + I2 -> (S4O6)2- + 2I^-

0.3克碘 = 0.3 / (126.9X2) = 1.182X10^-3
no. of mole of (S2O3)2- required = 1.182X10^-3 X 2 = 2.364 X 10^-3

2008-09-27 22:34:13 補充：
Volume of 0.1 M (S2O3)2- required = 2.364 X 10^-3 / 0.1 = 23.6 cm^3

2008-09-28 14:01:55 補充：
Sorry 上面有錯，由0.002 mole CuSO4.5H2O 產生了 0.001 mole I2 並未加入計算，
真實要處理的 I2 有 1.182X10^-3 + 0.001 = 2.182 X 10^-3 mole
要 0.1 M (S2O3)2- = 2X(2.182 X 10^-3) / 0.1 = 42.4 cm^-3