Question

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

Answer 1

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

Step-by-step 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* (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* (1mol\,S^(2-))/(1mol\,Na_(2)S)=2.50mol\,S^(2-)`