大连理工无机化学第四版课后题答案

340h4mu2022-10-04 11:39:541条回答

大连理工无机化学第四版课后题答案
18页思考题第4,7,10;
39页思考题5习题15;
69页习题10,18,19,20,21,22,23

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kaka120 共回答了18个问题 | 采纳率83.3%
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求助一道大连理工的无机化学题已知298K时E( MnO4- / Mn2+ ) = 1.51 V,E ( Fe3+ / F
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已知298K时E( MnO4- / Mn2+ ) = 1.51 V,E ( Fe3+ / Fe2+ ) = 0.771 V.用KMnO4溶液滴定FeSO4溶液,两者在酸性溶液中发生氧化还原反应.
(1) 写出相关反应的离子方程式,计算298K时该反应的标准平衡常数K ;
(2) 若开始时 c ( Fe2+ ) = 5.000 × 10-2 mol·L-1,c ( MnO4- ) = 1.000 × 10-2 mol·L-1,V ( Fe2+ ) = 20.00 mL,则反应达到平衡时,需要多少毫升KMnO4 溶液;
(3) 平衡时,若c ( H+ ) = 0.2000 mol·L-1,则 c ( Mn2+ )、c ( Fe3+ )、c ( MnO4- )、c ( Fe2+ )各为多少?
(4) 平衡时E( MnO4- / Mn2+ )、E ( Fe3+ / Fe2+ )各为多少?
(1) MnO4- + 5Fe2+ + 8H+ =Mn2+ + 5Fe2+ + 4H2O
K= 2.6 *1062
(2) c ( MnO4- ) V ( MnO4- ) = 1/5c ( Fe2+ ) V ( Fe2+ )
V ( MnO4- ) = 20mL
(3)设 平 衡 时 V(总)= 40.00 mL,c ’( MnO4- ) = x.
则 c ‘( Fe2+ ) = x,c ’( Mn2+ ) = 5.000 ×10-3 ,c ‘( Fe3+ ) = 2.500 ×10-2
为什么c ‘( Mn2+ ) = 5.000 ×10-3和设定V(总)= 40.00 mL?
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