1.998×10-3mol X mol
2Na + 2H2O ---> 2NaOH + H2↑
2 mol 1 mol
C1(OH-) = KB – С1(H+) = 10-14M2 / 10-9M = 10-5 M.
n1(OH-) = C1(OH-) × V = 10-5 M × 0.2 L = 2 × 10-6 mol.
C2(OH-) = KB – С2(H+) = 10-14M2 / 10-12M = 10-2 M.
n2(OH-) = C2(OH-) × V = 10-2 M × 0.2 L = 2 × 10-3 mol.
∆n(NaOH) = ∆n(OH-) = n2(OH-) – n1(OH-) = 2 × 10-3 mol – 2 × 10-6 mol = 1.998 × 10-3 mol.
n(H2) = X = 1.998×10-3 / 2 mol = 9.99×10-4 mol.
V(H2) = n(H2)×VM = 9.99×10-4 mol × 22.4 L/mol = 2.2378×10-2 L = 22.378 mL ≈ 22.38 mL.
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