Sodium thiosulfate formation
You are completely correct in your dimensional analysis and stoichiometry.


  36.8g is the correct answer.


Molar mass $\ce{Na2CO3 = 106g/mol }$

${25g \rightarrow 25/106 = 0.236}\text{ mol } \ce{Na2CO3}$

This will react with ${0.236*2 = 0.472} \text{ mol } \ce{Na2S}$

Molar mass $\ce{Na2S} = 78.0 \text{ g/mol}$

$0.472 \text{ mol } = 0.472*78.0 = 36.8 \text{ g } \ce{Na2S}$ will be needed.

I will now prove your book wrong

25.9g $\ce{Na2S}$ = 0.332mol $\ce{Na2S}$

0.332mol $\ce{Na2S}$ reacts with 0.166mol $\ce{Na2CO3}$

0.166mol $\ce{Na2CO3}$ = 17.5g of $\ce{Na2CO3}$, which is different from our starting quantity of 25.0g.