231k views
4 votes
How many moles of aqueous sodium ions and sulfide ions are formed when 2.05 mol of sodium sulfide dissolved in water

2 Answers

3 votes

Answer:

HEY ITS RIGHT ABOVE!!!

Step-by-step explanation:

i did 1.85 instead of 2.05 and it was right... im so sorry

User Daniel Hariri
by
4.2k points
2 votes

Step-by-step 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

User Nickmancol
by
5.5k points