Question

How many grams of sodium thiosulfate would be required to produce 64.3 g NaBr?

Answer 1

197.6 g

Step-by-step explanation:

Answer 2

This is an incomplete question, here is a complete question.

Sodium thiosulfate
`Na_2S_2O_3`, the major component in photographic fixer solution, reacts with silver bromide to dissolve it according to the following reaction:

`AgBr(s)+2Na_2S_2O_3(aq)\rightarrow Na_3Ag(S_2O_3)_2(aq)+NaBr(aq)`

How many grams of sodium thiosulfate would be required to produce 64.3 g NaBr?

Answer : The mass of sodium thiosulfate required would be, 197.6 grams.

Explanation : Given,

Mass of NaBr = 64.3 g

Molar mass of NaBr = 102.9 g/mol

Molar mass of
`Na_2S_2O_3` = 158.11 g/mol

The balanced chemical reaction is:

`AgBr(s)+2Na_2S_2O_3(aq)\rightarrow Na_3Ag(S_2O_3)_2(aq)+NaBr(aq)`

First we have to calculate the moles of
`NaBr`.

`\text{ Moles of }NaBr=\frac{\text{ Mass of }NaBr}{\text{ Molar mass of }NaBr}=(64.3g)/(102.9g/mole)=0.625moles`

Now we have to calculate the moles of
`Na_2S_2O_3`

From the reaction, we conclude that

As, 1 mole of
`NaBr` react to give 2 mole of
`Na_2S_2O_3`

So, 0.625 moles of
`NaBr` react to give
`0.625* 2=1.25` moles of
`Na_2S_2O_3`

Now we have to calculate the mass of
`Na_2S_2O_3`

`\text{ Mass of }Na_2S_2O_3=\text{ Moles of }Na_2S_2O_3* \text{ Molar mass of }Na_2S_2O_3`

`\text{ Mass of }Na_2S_2O_3=(1.25moles)* (158.11g/mole)=197.6g`

Thus, the mass of sodium thiosulfate required would be, 197.6 grams.