Given the following reaction Na2S2O3+AgBr+Na3{Ag(S2O3)2} How many moles if Na2S2O3 are needed to react completely with 42.7g of AgBr?

Assuming that the reaction you meant was:

#Na_2S_2O_3 + AgBr \rightarrow Na_3(Ag(S_2O_3)_2) + NaBr#

If I'm wrong tell me in the comments.

Step One: Write balanced equation

The reaction above is not balanced. The balanced equation would be:

#2Na_2S_2O_3 + AgBr \rightarrow Na_3(Ag(S_2O_3)_2) + NaBr#

Step Two: Find molar ratio between substances

Using the above we can see that the molar ratio of

#Na_2S_2O_3 : AgBr#

is equal to

#2 : 1#

(As the number in front of the substances represents the molar ratios)

Step Three: Find the number of moles of #AgBr#

Using the formula:

#n = m/M#

we can find out the number of moles of #AgBr#

#n = 42.7/(108 + 80)#

#n = 0.2274#

Step Four: Calculate the number of moles required

As mentioned before, the ratio between

#Na_2S_2O_3 : AgBr#

is equal to

#2 : 1#

Therefore, the number of moles of #Na_2S_2O_3# required is equal to:

#0.2274 \times 2 = 0.455#

Hope that makes sense!

#AgBr(s) + 2S_2O_3^(2-) rarr [Ag(S_2O_3)_2]^(3-)+Br^(-)#

#"Moles of silver bromide"=(42.7*g)/(187.77*g*mol^-1)=0.227*mol#..

And thus we require TWO EQUIV of sodium thiosulfate...i.e.

#0.227*molxx2xx158.11*g*mol^-1=??*g#