Let's solve this problem step-by-step using stoichiometry:
### Step 1: Calculate the number of moles of AgBr
We start by determining the number of moles of AgBr that we have. This can be calculated using the formula:
[tex]\[
\text{moles} = \frac{\text{mass}}{\text{molar mass}}
\][/tex]
Given:
- Mass of AgBr ([tex]\( \text{mass}_{\text{AgBr}} \)[/tex]) = 125.0 grams
- Molar mass of AgBr ([tex]\( M_{\text{AgBr}} \)[/tex]) = 187.70 grams per mol
[tex]\[
\text{moles}_{\text{AgBr}} = \frac{125.0 \, \text{g}}{187.70 \, \text{g/mol}} \approx 0.6659563 \, \text{mol}
\][/tex]
### Step 2: Use stoichiometry to find the moles of Ag2S2O3 formed
From the balanced chemical equation:
[tex]\[
2 \, \text{AgBr} + \text{Na}_2 \text{S}_2 \text{O}_3 \rightarrow \text{Ag}_2 \text{S}_2 \text{O}_3 + 2 \, \text{NaBr}
\][/tex]
We can see that 2 moles of AgBr produce 1 mole of Ag2S2O3. Therefore, the moles of Ag2S2O3 formed is half the number of moles of AgBr.
[tex]\[
\text{moles}_{\text{Ag}_2 \text{S}_2 \text{O}_3} = \frac{\text{moles}_{\text{AgBr}}}{2} \approx \frac{0.6659563 \, \text{mol}}{2} \approx 0.3329781 \, \text{mol}
\][/tex]
### Step 3: Calculate the mass of Ag2S2O3 formed
Now we'll use the number of moles of Ag2S2O3 to find the mass. The formula we use is:
[tex]\[
\text{mass} = \text{moles} \times \text{molar mass}
\][/tex]
Given:
- Moles of Ag2S2O3 ([tex]\( \text{moles}_{\text{Ag}_2 \text{S}_2 \text{O}_3} \)[/tex]) = 0.3329781 moles
- Molar mass of Ag2S2O3 ([tex]\( M_{\text{Ag}_2 \text{S}_2 \text{O}_3} \)[/tex]) = 327.74 grams per mol
[tex]\[
\text{mass}_{\text{Ag}_2 \text{S}_2 \text{O}_3} = 0.3329781 \, \text{mol} \times 327.74 \, \text{g/mol} \approx 109.13 \, \text{grams}
\][/tex]
So, the mass of [tex]\( \text{Ag}_2 \text{S}_2 \text{O}_3 \)[/tex] formed is approximately 109.13 grams.
### Summary
When 125.0 grams of AgBr reacts completely, approximately 109.13 grams of [tex]\( \text{Ag}_2 \text{S}_2 \text{O}_3 \)[/tex] are formed.
