There are 1.16x10²² sulfide ions (S²⁻) in 15 dg of sodium sulfide (Na₂S).
The equation for sodium sulfide is the following:
Na₂S → 2Na⁺ + S²⁻ (1)
So from equation (1), we can see that in 1 mol of sodium sulfide we have 1 mol of sulfide ions and 2 moles of sodium ions.
First, let's find the number of moles of Na₂S
[tex] n_{Na_{2}S} = \frac{m_{Na_{2}S}}{M_{Na_{2}S}} [/tex] (2)
Where:
[tex]m_{Na_{2}S}[/tex]: is the mass of Na₂S = 15 dg = 1.5 g
[tex]M_{Na_{2}S}[/tex]: is the molar mass of Na₂S = 78.0452 g/mol
The number of moles of Na₂S is (eq 2):
[tex] n_{Na_{2}S} = \frac{1.5 g}{78.0452 g/mol} = 0.0192 \:mol [/tex]
We can find the number of ions of S²⁻ with Avogadro's number, knowing that the number of moles of Na₂S is equal to the number of moles of S²⁻ (eq 1).
[tex] ions_{S^{2-}} = \frac{6.022 \cdot 10^{23} \:ions}{1\:mol}*0.0192 \:mol = 1.16 \cdot 10^{22} \:ions [/tex]
Therefore, there are 1.16x10²² sulfide ions in 15 dg of sodium sulfide.
Find more about Avogadro's number here:
I hope it helps you!
