因为h2s为弱酸，所以要看h2s的量： h2s不足：h2s+2naoh=na2s+2h2o; h2s过量：h2s+naoh=nahs+h2o. 离子方程式拆开就行了

少量h2s：2naoh + h2s = na2s + 2h2o （1） 过量h2s：naoh + h2s = nahs + h2o （2） （2）式就是让（1）式得到的产物na2s再和h2s反应： na2s + h2s = 2nahs （3） （1）+（3）式得到（2）

H2S + NaOH = Na2S + H2O is a Double Displacement (Acid-Base) reaction where one mole of solid Hydrogen Sulfide [H 2 S] and two moles of aqueous Sodium Hydroxide [NaOH] react to form one mole of aqueous Sodium Sulfide [Na 2 S] and two moles of liquid Water [H 2 O]

1 h 2 s + 2 naoh = 1 na 2 s + 2 h 2 o H is balanced: 4 atoms in reagents and 4 atoms in products. All atoms are now balanced and the whole equation is fully balanced:

naoh过量（或h2s少量）,反应方程式是2naoh+h2s=na2s+2h2o； naoh少量（或h2s过量）,会有两种生成物,即na2s和nahs,反应式也就有两个,分别是 naoh+h2s＝nahs+h2o和2naoh+h2s＝na2s+2h2o

书写naoh溶液与h2s 反应（h2s 少）的离子方程式. 百度试题 结果1. 题目. 书写naoh溶液与h2s 反应（h2s 少）的离子方程式 . 相关知识点： 试题来源： 解析. 氢氧化钠与少量的硫化氢反应生成硫化钠和水，离子方程式为：h2s+2oh-=s2-+2h2o . ...

NaOH + H2S = Na2S + H2O is a Double Displacement (Acid-Base) reaction where two moles of aqueous Sodium Hydroxide [NaOH] and one mole of solid Hydrogen Sulfide [H 2 S] react to form one mole of aqueous Sodium Sulfide [Na 2 S] and two moles of liquid Water [H 2 O]

这个问题属于碱和酸性气体反应的例子，这个反应的 化学方程式 为：H2S+2NaOH = Na2S+H2O (H2S足量)H2S+NaOH = NaHS+H2O (H2S过量)这有个规律，就是碱和这样的气体反应（CO2、SO2 、SO3)等，当气体不足的时候生成 正盐，当气体足量的时候生成 酸式盐。 例如上面的例子就是。 但是有个特例就是苯酚盐与这样的气体. 反应只生成酸式盐。 明白不？ 不明白可以随时问我。

1 naoh + 1 h 2 s = 1 h 2 o + 1 na 2 s For each element, we check if the number of atoms is balanced on both sides of the equation. Na is not balanced: 1 atom in reagents and 2 atoms in products.

Solved and balanced chemical equation H2S + NaOH → NaHS + H2O with completed products. Application for completing products and balancing equations.