How many moles of aqueous sodium ions and sulfide ions are formed when 2.05 mol of sodium sulfide dissolved in water

Explanation:

We have to put out the balanced chemical equation before proceeding. The equation for the dissociation of sodium sulfide in water is given as;

Na₂S  --> 2Na⁺ + S²⁻

From the stoichiometry of the reaction we can tell that;

1 mol of Na₂S produces 2 moles of Na⁺ and 1 mol of S²⁻.

Sodium ion:

1 mol of Na₂S  = 2 mol of Na⁺

2,05 = x

upon solving for x;

x =  2 * 2.05 = 4.10 moles

Sulfide ion:

1 mol of Na₂S  = 1 mol of S²⁻

2.05 = x

upon solving for x;

x = 1 * 2.05 = 2.05 moles

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