Sodium thiosulfate formation

I have a problem solivng this chemical problem. Any help would be highly appreciative.


Sodium Thiosulfate can be made by the reaction


$\ce{Na\_2CO\_3 + 2Na\_2S + 4SO\_2 -> 3Na\_2S\_2O\_3 + CO\_2}$


Question: How many grams of $\ce{Na\_2S}$ are required to react with 25.0 g of $\ce{Na\_2CO\_3}$ if $\ce{SO\_2}$ is present in excess?


My attempt: 


$25.0 \text{ g } \ce{Na\_2CO\_3}\frac{1 \text{ mol } \ce{Na\_2CO\_3}}{105.99}\frac{2}{1}\frac{78.05}{1}$


The answer from the book states: 25.9g$\ce{Na\_2S}$




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.


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