两个题 就是氧化还原那块的~第一个An organic compound contains carbon,hydroge

两个题
就是氧化还原那块的~第一个An organic compound contains carbon,hydrogen,and sulfur.A sample of it with a mass of 1.886 g was burned in oxygen to give gaseous CO2,H2O,and SO2.These gases were passed through 313.3 mL of an acidified 0.0200 M KMnO4 solution,which caused the SO2 to be oxidized to SO42-.Only part of the available KMnO4 was reduced to Mn2+.Next,31.33 mL of 0.0300 M SnCl2 was added to 31.33 a mL portion of this solution,which still contained unreduced KMnO4.There was more than enough added SnCl2 to cause all of the remaining MnO4- in the 31.33 mL portion to be reduced to Mn2+.The excess Sn2+ that still remained after the reaction was then titrated with 0.0100 M KMnO4,requiring 0.02242 L of the KMnO4 solution to reach the end point.Based upon all this data,the the percentage of sulfur in the original sample of the organic compound that had been burned is %.
第二个Solder is an alloy containing the metals tin and lead.A particular sample of this alloy weighing 1.50 g was dissolved in acid.All the tin was then converted to the +2 oxidation state.Next,it was found that 0.368 g of Na2Cr2O7 was required to oxidize the Sn2+ to Sn4+ in an acidic solution.In the reaction the chromium was reduced to Cr3+ ion.
Write a balanced net ionic equation for the reaction between Sn2+ and Cr2O72- in an acidic solution.

1.题目翻译：
①某有机物含有C、H、S元素.质量为1.886g的样品在氧气中燃烧,生成CO2,H2O和SO2.
②这些气体被通入313.3ml浓度为0.0200mol/L的酸性KMnO4溶液,使得SO2被氧化为SO4 2-,同时一部分KMnO4被还原为Mn 2+.
③然后将向31.33ml的剩余溶液样品（仍含有未被还原的KMnO4）中加入31.33ml浓度为0.0300mol/L的SnCl2溶液,其中的SnCl2为过量,足以将剩余的KMnO4全部还原为Mn 2+.
④反应后用0.0100mol/L的KMnO4溶液滴定测量溶液中剩余的SnCl2,使用0.02242L的该KMnO4溶液恰好达到滴定终点.
根据上述条件,求被燃烧的有机物样品中S元素的质量分数.
依次发生反应：
①CxHySz+(x+0.25y+z)O2→xCO2+(0.5y)H2O+zSO2
②5SO2+2(MnO4 -)+2H2O=2(Mn 2+)+5(SO4 2-)+4(H +)
③2(MnO4 -)+5(Sn 2+)+16(H +)=2(Mn 2+)+5(Sn 4+)+8H2O
④2(MnO4 -)+5(Sn 2+)+16(H +)=2(Mn 2+)+5(Sn 4+)+8H2O
其中③④反应一样,其中③是在使用SnCl2还原剩余的KMnO4时发生的,④是在使用KMnO4滴定剩余的SnCl2时发生的.
由于题干按照实验顺序进行,而且求初始量,因此用逆推法.
最后滴定恰好完成时,使用KMnO4物质的量为0.02242×0.01=0.0002242mol
由反应④可知被滴定的剩余SnCl2物质的量为0.0005605mol
而此前加入的SnCl2总物质的量为0.03×0.03133=0.0009399mol≈0.00094mol
因此在③中反应掉的SnCl2物质的量为0.00094-0.0005605=0.0003794mol≈0.00038mol
由③可知参加反应的KMnO4物质的量为0.00015176mol≈0.000152mol
又因为第③步只取了第②步剩余溶液的10%,因此在第②步过后剩余的KMnO4总物质的量应为0.00015176×10=0.0015176mol≈0.00152mol
又因为最初KMnO4总物质的量为0.3133×0.02=0.006266mol
因此在①中反应掉的KMnO4物质的量为0.006266-0.0015176=0.0047484mol
由反应①可知,被氧化的SO2物质的量为0.011871mol,因此S原子也为0.011871mol,质量为0.379872g
因此S的质量分数为0.379872/1.886×100%≈20.14%
2.题目翻译：
焊锡是一种含有锡和铅的合金.取1.50g的该合金样品溶于酸中,其中的锡都转化为+2价氧化态（即变为Sn 2+离子）.然后发现,需要0.368g的Na2Cr2O7将Sn 2+全部氧化为Sn4+；在该反应中,Cr元素被还原为Cr 3+离子.要求写出配平后的在酸性溶液中,Sn 2+与Cr2O7 2-反应的离子方程式.
3(Sn 2+)+(Cr2O7 2-)+14(H +)=3(Sn 4+)+2(Cr 3+)+7H2O
我很困惑如果只要写一个方程式的话为什么题目中要给出具体数据……我原以为它应该是会要答题者求出合金中Sn的质量分数.如果要求这个的话,因为已知Na2Cr2O7的质量为0.368g,可算出其物质的量,再由方程式可求Sn 2+物质的量；由于原合金中的Sn单质全部转化为离子,因此再求出Sn元素的质量即可.
数字有一些麻烦,我的思路是没错的,但可能会出现一点运算错误,建议你可以再检查一下.BTW,难道这就是传说中的SAT II的考题么?