S8(s) + Na2SO3(aq) + H2O(l) Na2S2O3·5 H2O(s) HELP?!?

Question

The compound sodium thiosulfate pentahydrate, Na2S2O3·5 H2O, is important commercially to the photography business as "hypo," because it has the ability to dissolve unreacted silver salts from photographic film during development. Sodium thiosulfate pentahydrate can be produced by boiling elemental sulfur in an aqueous solution of sodium sulfite.
S8(s) + Na2SO3(aq) + H2O(l) Na2S2O3·5 H2O(s) (unbalanced)
What is the theoretical yield of sodium thiosulfate pentahydrate when 3.18 g of sulfur is boiled with 14.9 g of sodium sulfite?
........ g?

Sodium thiosulfate pentahydrate is very soluble in water. What is the percent yield of the synthesis if a student doing this experiment is able to isolate (collect) only 2.7 g of the product?
.......... %?

THANK YOU
please explain how you got the answer because I would like to learn how to do this!

Answer 1

S8(s) + Na2SO3(aq) + H2O(l) Na2S2O3·5 H2O(s) HELP?!?

The compound sodium thiosulfate pentahydrate, Na2S2O3·5 H2O, is important commercially to the photography business as "hypo," because it has the ability to dissolve unreacted silver salts from photographic film during development. Sodium thiosulfate pentahydrate can be produced by boiling elemental sulfur in an aqueous solution of sodium sulfite.

S8(s) + Na2SO3(aq) + H2O(l) Na2S2O3·5 H2O(s) (unbalanced)

What is the theoretical yield of sodium thiosulfate pentahydrate when 3.18 g of sulfur is boiled with 14.9 g of sodium sulfite?

1] This is a limiting reagent problem

2] Calculate the number of moles in 3.18 g of sulfur is boiled and in 14.9 g of sodium sulfite

3] Balance the reaction:

S8(s) + 8 Na2SO3(aq) + 40 H2O(l) --->8 Na2S2O3·5 H2O(s)

4] Now multiply the moles of S8 by 8. Is that less than the moles of Na2SO3 If so, S8 limits. If that is greater than the moles of Na2SO3, then Na2SO3 limits. If it is equal, then neither [and both] limits.

Take the number of moles of the limiting reagent and use the coefficients of the balanced reaction to tell you how many moles of product you will get, then convert that to grams.



Sodium thiosulfate pentahydrate is very soluble in water. What is the percent yield of the synthesis if a student doing this experiment is able to isolate (collect) only 2.7 g of the product?

.......... %?

[Grams he got/theoretical yield in grams] x 100

Answer 2

The balanced equation is 5 S8 + 40 Na2SO3 + 8 H2O = 8 (Na2S2O3)5H2O



1) S8: 3.18 grams / (8 x 32.065) the MW of S8 = 0.0124 mols S8

divide by 5 to determine quantity of (5 S8) to determine fit for the above balanced equation which is ** 0.00248 **

2) Na2SO3: 14.9 grams / 126.043 the MW of sodium sulfite = 0.1182 mols Na2SO3

divide by 40 to determine quantity of (40 Na2SO3) to determine fit for the above balanced equation which is ** 0.002955 ** (higher than S8 value so an excess of this reactant is available.

3) Assuming an excess of water (since no limit was given), the quantity of sulfur will limit the amount of product formed. (**0.00248**) (8 mols product)(808.4 the MW of (Na2S2O3)5H2O) = 16.04 grams *** This is the theoretical amount opposite the 3.18 grams S8 reactant

4) check using material balance: grams of reactants = grams of product

a) S8: 3.18 given

b) Na2SO3: 0.00248 x 40 x 126.043 = 12.5

c) Water: 0.00248 x 8 x 18.02 = 0.36

"Total"; 3.18 + 12.5 + 0.36 = 16.04 grams QED





Percent Yield = Actual / Theo x 100 = 2.7/16.04 x 100 = 16.8 %

Answer 3

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