This is an incomplete question, here is a complete question.
Sodium thiosulfate
, the major component in photographic fixer solution, reacts with silver bromide to dissolve it according to the following reaction:
![AgBr(s)+2Na_2S_2O_3(aq)\rightarrow Na_3Ag(S_2O_3)_2(aq)+NaBr(aq)](https://tex.z-dn.net/?f=AgBr%28s%29%2B2Na_2S_2O_3%28aq%29%5Crightarrow%20Na_3Ag%28S_2O_3%29_2%28aq%29%2BNaBr%28aq%29)
How many grams of sodium thiosulfate would be required to produce 64.3 g NaBr?
Answer : The mass of sodium thiosulfate required would be, 197.6 grams.
Explanation : Given,
Mass of NaBr = 64.3 g
Molar mass of NaBr = 102.9 g/mol
Molar mass of
= 158.11 g/mol
The balanced chemical reaction is:
![AgBr(s)+2Na_2S_2O_3(aq)\rightarrow Na_3Ag(S_2O_3)_2(aq)+NaBr(aq)](https://tex.z-dn.net/?f=AgBr%28s%29%2B2Na_2S_2O_3%28aq%29%5Crightarrow%20Na_3Ag%28S_2O_3%29_2%28aq%29%2BNaBr%28aq%29)
First we have to calculate the moles of
.
![\text{ Moles of }NaBr=\frac{\text{ Mass of }NaBr}{\text{ Molar mass of }NaBr}=\frac{64.3g}{102.9g/mole}=0.625moles](https://tex.z-dn.net/?f=%5Ctext%7B%20Moles%20of%20%7DNaBr%3D%5Cfrac%7B%5Ctext%7B%20Mass%20of%20%7DNaBr%7D%7B%5Ctext%7B%20Molar%20mass%20of%20%7DNaBr%7D%3D%5Cfrac%7B64.3g%7D%7B102.9g%2Fmole%7D%3D0.625moles)
Now we have to calculate the moles of ![Na_2S_2O_3](https://tex.z-dn.net/?f=Na_2S_2O_3)
From the reaction, we conclude that
As, 1 mole of
react to give 2 mole of ![Na_2S_2O_3](https://tex.z-dn.net/?f=Na_2S_2O_3)
So, 0.625 moles of
react to give
moles of ![Na_2S_2O_3](https://tex.z-dn.net/?f=Na_2S_2O_3)
Now we have to calculate the mass of ![Na_2S_2O_3](https://tex.z-dn.net/?f=Na_2S_2O_3)
![\text{ Mass of }Na_2S_2O_3=\text{ Moles of }Na_2S_2O_3\times \text{ Molar mass of }Na_2S_2O_3](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DNa_2S_2O_3%3D%5Ctext%7B%20Moles%20of%20%7DNa_2S_2O_3%5Ctimes%20%5Ctext%7B%20Molar%20mass%20of%20%7DNa_2S_2O_3)
![\text{ Mass of }Na_2S_2O_3=(1.25moles)\times (158.11g/mole)=197.6g](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DNa_2S_2O_3%3D%281.25moles%29%5Ctimes%20%28158.11g%2Fmole%29%3D197.6g)
Thus, the mass of sodium thiosulfate required would be, 197.6 grams.