The following reaction is a key step in black-and-white photography (see p. 1035):

$\mathbf{AgBr}\left(s\right)\mathbf{+}\mathbf{2}{\mathbf{S}}_{\mathbf{2}}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{2}\mathbf{-}}\left(\mathrm{aq}.\right)\mathbf{\to }\mathbf{Ag}{{\left(S_2{O}_1\right)}_{\mathbf{2}}}^{\mathbf{3}\mathbf{-}}\left(\mathrm{aq}.\right)\mathbf{+}{\mathbf{Br}}^{\mathbf{-}}\left(\mathrm{aq}.\right)$

During fixing,$\mathbf{258}\mathbf{}\mathbf{mL}$of hypo (sodium thiosulfate) was used. The hypo concentration was 0.1052M before the AgBr reacted, and 0.0378M afterward. How many grams ofreacted?

258 mL of sodium thiosulphate solution is converted from 0.1052M to 0.0378M.

A coordination complex consists of a central atom or ion, which is typically metallic and is named the coordination center, and a surrounding array of bound molecules or ions, that are successively called ligands or complexing agents.

Get the amount of thiosulphate in moles by subtracting the final morality from the initial molarity and then multiplying it by the volume given. This amount is the amount of thiosulphate that reacted according to this equation:

$\mathrm{AgBr}\left(\mathrm{s}\right)+2{\mathrm{S}}_2{{\mathrm{O}}_3}^{2-}\to \mathrm{Ag}{{\left({\mathrm{S}}_2{\mathrm{O}}_3\right)}_3}^{3-}\left(\mathrm{aq}.\right)+{\mathrm{Br}}^{-}\left(\mathrm{aq}.\right)$

$$\begin{gathered} 0.1052\mathrm{M}-0.0378\mathrm{M}=0.0674\mathrm{M} \\ \mathrm{From}\mathrm{the}\mathrm{molar}\mathrm{calculation},\mathrm{we}\mathrm{can}\mathrm{write} \\ 0.0647\mathrm{M}\times \frac{258\mathrm{mL}}{1000\mathrm{mL}}=0.017389\mathrm{mol}\mathrm{of}{\mathrm{S}}_2{{\mathrm{O}}_3}^{2-} \end{gathered}$$

Convert the moles of ${\mathrm{S}}_2{{\mathrm{O}}_3}^{2-}$to AgBr by multiplying it according to the molar ratio of 1 AgBr: 2${\mathrm{S}}_2{{\mathrm{O}}_3}^{2-}$(as seen in the reaction). This is the amount of AgBr that reached in moles.

$0.017389\mathrm{mol}{\mathrm{S}}_2{{\mathrm{O}}_3}^{2-}\times \frac{1\mathrm{mol}\mathrm{AgBr}}{2\mathrm{mol}{\mathrm{S}}_2{{\mathrm{O}}_3}^{2-}}=0.0086946\mathrm{mol}\mathrm{AgBr}$

Convert the moles of AgBr into grams using the molecular weight of AgBr, which is 187.77 gm fro 1 mole.

Therefore the mass of 0.086946 mol of AgBr will be,

$0.086946\mathrm{mol}\times 187.77\mathrm{gm}=1.633258\mathrm{gm}$