What mass of Mo2O3 (MW 240.0g/mol) was present in the sample?
(3MnO4-)+ (5Mo+3) + (8H2O) ---> (3Mn+2) + (5MoO4-2) +(16H+)
It takes 10.0 mL of 0.100 M solution of KMnO4 to completely and exactly react all the Mo+3 in a sample of Mo2O3 following the reaction below?
2015-09-01 6:22 pm
回答 (2)
2015-09-01 8:34 pm
✔ 最佳答案
3 MnO4{-} + 5 Mo{3+} + 8 H2O → 3 Mn{2+} + 5 MoO4{2-} + 16 H{+}(0.0100 L) x (0.100 mol/L KMnO4) x (1 mol MnO4{-} / 1 mol KMnO4) x (5 mol Mo{3+} / 3 mol MnO4{-}) x
(1 mol Mo2O3 / 2 mol Mo{3+}) x (240.0 g Mo2O3/mol) = 0.200 g Mo2O3
2015-09-01 10:13 pm
3MnO₄⁻+ 5Mo³⁺ + 8H₂O → 3Mn²⁺ + 5MoO₄⁻ +16H⁺
No. of moles of KMnO₄ = 0.100 × (10.0/1000) = 0.001 mol
No. of moles of MnO₄⁻ = 0.001 mol
No. of moles of Mo³⁺ = 0.001 × 5/3 = 0.00167 mol
No. of moles of Mo₂O₃ = 0.00167 × (1/2) = 0.000835 mol
Mass of Mo₂O₃ = 0.000835 × 240 = 0.200 g
No. of moles of KMnO₄ = 0.100 × (10.0/1000) = 0.001 mol
No. of moles of MnO₄⁻ = 0.001 mol
No. of moles of Mo³⁺ = 0.001 × 5/3 = 0.00167 mol
No. of moles of Mo₂O₃ = 0.00167 × (1/2) = 0.000835 mol
Mass of Mo₂O₃ = 0.000835 × 240 = 0.200 g
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