201 grams.
Explanation
What's the formula of Silver Sulfide?
Consider the two kinds of ions in silver sulfide:
- Silver ions [tex]\text{Ag}^{+}[/tex], and
- Sulfide ions [tex]\text{S}^{2-}[/tex].
The charge on the two types of ions should balance. The charge on each sulfide ion is twice as much as that on a silver ion. There needs to be two silver ions for each sulfide ion.
Hence the formula: [tex]\text{Ag}_{\bf 2} \text{S}[/tex].
How many moles of formula units in 174 grams of Ag₂S.
Relative atomic mass from a modern periodic table:
- Silver Ag: 107.868;
- Sulfur S: 32.06;
- Chlorine Cl: 35.45.
[tex]M(\text{Ag}_{2}\text{S}) = 2 \times 107.868 + 32.06 = 247.796 \; \text{g} \cdot \text{mol}^{-1}[/tex].
[tex]n = \dfrac{m}{M} = \dfrac{174}{247.796} = 0.702 \; \text{mol}[/tex].
How many moles of AgCl formula units will be formed from 174 grams of Ag₂S?
The question doesn't mention the other product in this reaction. Finding that product isn't a must as long as all Ag from Ag₂S end up in AgCl.
[tex]1\; \text{Ag}_{2}\text{S} + ?\; \text{HCl} \to ? \; \text{AgCl} + ? \; \text{H}_2\text{S}[/tex].
We are trying to answer the question: how many moles of AgCl does each mole formula unit of Ag₂S produce. Thus, assign a coefficient of one to Ag₂S.
There are two silver atoms in each formula unit Ag₂S. Both would end up in AgCl. However, there's only one silver atom in each formula unit of AgCl. Two Ag atoms would make two formula units of AgCl. In other words, each mole formula units of Ag₂S would produce two moles formula units of AgCl.
[tex]1\; \text{Ag}_{2}\text{S} + ?\; \text{HCl} \to {\bf 2} \; \text{AgCl} + ? \; \text{H}_2\text{S} [/tex].
There are 0.702 moles of Ag₂S formula units in 174 grams of the compound. That would lead to [tex]2 \times 0.702 = 1.40 \; \text{mol}[/tex] of AgCl.
What's the mass of 1.40 mol of AgCl?
[tex]M(\text{AgCl}) = 107.868 + 35.45 = 143.318 \; \text{g} \cdot \text{mol}^{-1}[/tex].
[tex]m = n \cdot M = 1.40 \times 143.318 = 201 \; \text{g}[/tex].
