Question

Given the following reaction: Na2S2O3  + AgBr (yields) NaBr + Na3 [Ag(S2O3)2 ]

a. How many moles of Na2S2O3  are needed to react completely with 42.7 g of
AgBr?
b. What is the mass of NaBr that will be produced from 42.7 g of AgBr?

I need you to show the work please.

Answer 1

Na2S2O3 + AgBr → NaBr + Na3[Ag(S2O3)2]

Step-by-step explanation:

First check that the given equation is balanced ... it isn't ... so the first thing to do is balance the equation:

balancing Na:

2Na2S2O3 + AgBr → NaBr + Na3[Ag(S2O3)2]

and everything is now balanced so we've got the balanced equation

molar mass AgBr = 107.87 + 79.90 = 187.77 g/mol

mol AgBr available = 42.7 g AgBr x [1 mol / 187.77 g] = 0.2274 mol AgBr

from the balanced equation the mole ratio AgBr : Na2S2O3 = 1 : 2

so mol Na2S2O3 required = 0.2274 mol AgBr x [ 2 mol Na2S2O3 / mol AgBr]

= 0.455 mol Na2S2O3 (to 3 sig figs)