Question

How many moles of aqueous sodium ions and sulfide are formed when 2.50 mil of sodium sulfide dissolved in water.

Answer 1

Answer:

5.00 and 2.50 moles of aqueous sodium and sulfide ions are formed.

Explanation:

The dissociation reaction of sodium sulfide is as follows.

$Na_{2}S(aq)\rightarrow 2Na^{+}(aq)+S^{2-}(aq)$

From the reaction one mole of sodium sulfide produce 2 moles sodium ions.

Let's calculate the moles of $Na^{+}$ions.

$Moles\,of\,Na^{+}=2.50molNa_{2}S\times \frac{2mol\,Na^{+}}{1mol\,Na_{2}S}=5.00mol\,Na^{+}$

From the reaction one mole of sodium sulfide produce one mole of sulfide ions.

Let's calculate the moles of $S^{2-}$ions.

$Moles\,of\,S^{2-}=2.50molNa_{2}S\times \frac{1mol\,S^{2-}}{1mol\,Na_{2}S}=2.50mol\,S^{2-}$