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 3.21 g of sulfur is boiled with 15.7 g of sodium sulfite?


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 11.7 g of the product?

2.

2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)

If 4.99 g of potassium chlorate is heated, what theoretical mass of oxygen gas should be produced?

If only 1.10 g of oxygen is actually obtained, what is the percent yield?

please help !!

Answer 1

1)

we have to know the balanced chemical reaction

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

so 1 mol of S8 reacts with 8 mol of  Na2SO3 to give 8 mol of Na2S2O3·5 H2O

256.52 gm of S8 reacts with 8 * 126.04 gm of Na2SO3 to give 8 * 158.10 gm of Na2S2O3·5 H2O

In this reaction the limiting reagent is the S8 which is 3.21 gm

so 3.21 gm of S8 will react = 8 * 126.04 / 256.52 * 3.21 = 12.61 gm of Na2SO3

256.52 gm of S8 gives = 1264.8 gm of  Na2S2O3·5 H2O

so 3.21 gm of S8 will give = 1264.8 / 256.52 * 3.21 = 15.82 gm of Na2S2O3·5 H2O

So theoritical yield = 15.82 gm

Actual yield = 11.7 gm,

So % yield = 11.7 /15.82 *100 = 73.95 %

2)

2 KClO3(s) → 2 KCl(s) + 3 O2(g)

2 mol of KClO3 produces 2 mol of KCl and 3 mol of O2

2 * 122.54 gm of KClO3 produces 3 * 32 gm of O2

so 4.99 gm of KClO3 will produce = 1.954 gm of O2

So theoritical yield = 1.954 gm

Actual yield = 1.10 gm

So % yield = 1.10 / 1.954 * 100 = 56.29 %