Question

Silver bromide can be "dissolved" by the action of aqueous Na₂S₂O₃ (called "hypo") after any silver that has been "activated" by exposure to light is reduced to metallic silver by another reagent. This process is the basis for the development of black-and-white film. Calculate the mass of hypo necessary to dissolve 42.6 g of AgBr. The equation for the dissolving process is given below.

AgBr(s) + 2 Na₂S₂O₃(aq)  Na₃Ag(S₂O₃)₂(aq) + NaBr(aq)

Answer 1

Ans:-

We have given that the equation is $AgBr(s) + 2 Na_2S_2O_3 (aq)\rightarrow Na_3Ag(S_2O_3)_2(aq) + NaBr(aq)$

So firstly we find the moles of $AgBr$

mol of $AgBr=\dfrac{42.6}{187.77}=0.22mol$

So according to the stoichiometry $1mol$ of $AgBr$ dissolve with $2 mol$ of Hypo solution$Na_2S_2O_3 (aq)$

Hence mol of Hypo($Na_2S_2O_3 (aq)$ ) solution is$0.44mol$

The mass of Hypo solution $= 0.44\times158.11=69.56g$