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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$\ce{Na_2S}$

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  • $\begingroup$ Welcome to Chem.SE! Its good to see proper formatting of chemical equations in one's first post. Are you missing an equality sign under My attempt ? $\endgroup$ – Del Pate Mar 29 '15 at 5:34
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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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  • $\begingroup$ The answer from the book says 25.9g Na_2(S) $\endgroup$ – user130431 Mar 29 '15 at 3:18
  • $\begingroup$ I'm certain your book is wrong. We both solved in different layouts and got the same answer. I'll check and see what may be the issue. $\endgroup$ – Nerdatope Mar 29 '15 at 3:23

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