Question

how many sulfide ions are present in 23.5g of Sb2S3

Answer 1

You can figure it out by doing dimensional analysis and plugging in the information you're given and then using the answer to your dimensional analysis to do percent composition and then your answer to that will be your final answer.

Answer 2

Answer: The number of sulfide ion are, $1.25\times 10^{23}$

Explanation : Given,

Mass of $Sb_2S_3$ = 23.5 g

Molar mass of $Sb_2S_3$ = 339.7 g/mol

First we have to calculate the moles of $Sb_2S_3$

$\text{Moles of }Sb_2S_3=\frac{\text{Given mass }Sb_2S_3}{\text{Molar mass }Sb_2S_3}$

$\text{Moles of }Sb_2S_3=\frac{23.5g}{339.7g/mol}=0.0692mol$

Now we have to calculate the moles of sulfide ion.

In $Sb_2S_3$, there are 2 moles of $Sb^{3+}$ ion and 3 moles of $S^{2-}$ ion.

As, 3 mole of sulfide ion contains $3\times 6.022\times 10^{23}$ sulfide ion

So, 0.0692 mole of sulfide ion contains $0.0692\times 3\times 6.022\times 10^{23}=1.25\times 10^{23}$ sulfide ion

Thus, the number of sulfide ion are, $1.25\times 10^{23}$