(1)①反应中N元素被还原性,HNO
3为氧化剂,故答案为:HNO
3;
②反应的离子方程式为Cu+4H
++2NO
3-=Cu
2++2NO
2↑+2H
2O,故答案为:Cu+4H
++2NO
3-=Cu
2++2NO
2↑+2H
2O;
(2)①反应中H
2S中S氧化化合价升高,由-2价升高到0价,SO
2中S元素化合价降低,由+4价降低到0价,氧化还原反应反应中得失电子数目相等,表现为化合价升降的总数相等,为
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/279759ee3d6d55fbd28391976e224f4a20a4dd3e?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
,故答案为:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/279759ee3d6d55fbd28391976e224f4a20a4dd3e?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;
②反应2H
2S+SO
2═3S↓+2H
2O中,生成3molS,氧化产物为2mol,生成产物为1mol,转移电子4mol,若氧化产物比还原产物多1.6g,即多0.05mol,则转移电子0.2mol,
电子数为0.2N
A,故答案为:0.2N
A.