已知N2H4中氢为+1价.2N2H4+N2H4=3N2+4H2O,已知N2H4和N2H4总物质的量为6mol,还原产物比氧化产物少42g
问题描述:
已知N2H4中氢为+1价.2N2H4+N2H4=3N2+4H2O,已知N2H4和N2H4总物质的量为6mol,还原产物比氧化产物少42g
,选(选正确的A,转移8mol电子
B,原混合物中含3molN2H4
c,原混合物中含6mol N
D,还原产物N2为1.5mol
ABCD每个答案分别应该为多少?B为什么不对
答
应该选择D 应该是:2N2H4+N2O4=3N2+4H2O上式可改为:2N2H4+N2O4=2N2(被氧化) +N2 (被还原)+ 4H2O ----转移8e- 2121 8 2x...