Question

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?

Answer 1

`0.455` moles

Explanation:

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!

Answer 2

We examine the appropriate reaction....and get a mass of approx. `72*g` with respect to sodium thiosulfate.....

Explanation:

`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`