This is an incomplete question, here is a complete question.
Sodium thiosulfate [tex]Na_2S_2O_3[/tex], the major component in photographic fixer solution, reacts with silver bromide to dissolve it according to the following reaction:
[tex]AgBr(s)+2Na_2S_2O_3(aq)\rightarrow Na_3Ag(S_2O_3)_2(aq)+NaBr(aq)[/tex]
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 [tex]Na_2S_2O_3[/tex] = 158.11 g/mol
The balanced chemical reaction is:
[tex]AgBr(s)+2Na_2S_2O_3(aq)\rightarrow Na_3Ag(S_2O_3)_2(aq)+NaBr(aq)[/tex]
First we have to calculate the moles of [tex]NaBr[/tex].
[tex]\text{ Moles of }NaBr=\frac{\text{ Mass of }NaBr}{\text{ Molar mass of }NaBr}=\frac{64.3g}{102.9g/mole}=0.625moles[/tex]
Now we have to calculate the moles of [tex]Na_2S_2O_3[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]NaBr[/tex] react to give 2 mole of [tex]Na_2S_2O_3[/tex]
So, 0.625 moles of [tex]NaBr[/tex] react to give [tex]0.625\times 2=1.25[/tex] moles of [tex]Na_2S_2O_3[/tex]
Now we have to calculate the mass of [tex]Na_2S_2O_3[/tex]
[tex]\text{ Mass of }Na_2S_2O_3=\text{ Moles of }Na_2S_2O_3\times \text{ Molar mass of }Na_2S_2O_3[/tex]
[tex]\text{ Mass of }Na_2S_2O_3=(1.25moles)\times (158.11g/mole)=197.6g[/tex]
Thus, the mass of sodium thiosulfate required would be, 197.6 grams.
