Photographic film contains silver bromide in gelatin. Once exposed, some of the silver bromide decomposes, producing fine grains of silver. The unexposed silver bromide is removed by treating the film with sodium thiosulfate. Soluble sodium silver thiosulfate (Na3Ag(S2O3)2) is produced.

Question

Photographic film contains silver bromide in gelatin. Once exposed, some of the silver bromide decomposes, producing fine grains of silver. The unexposed silver bromide is removed by treating the film with sodium thiosulfate. Soluble sodium silver thiosulfate (Na3Ag(S2O3)2) is produced.

AgBr(s) + 2 Na2S2O3(aq) Na3Ag(S2O3)2(aq) + NaBr(aq)
Determine the mass of Na3Ag(S2O3)2 produced if 0.275 g AgBr is removed.

Answer 1

Here is a worked example of a stoichiometry problem such as yours.

http://www.jiskha.com/science/chemistry/stoichiometry.html Just follow the steps.

Answer 2

.588 g Na3Ag(S2O3)2

Answer 3

To determine the mass of Na3Ag(S2O3)2 produced if 0.275 g AgBr is removed, we will use stoichiometry.

Given:
Mass of AgBr = 0.275 g

The balanced chemical equation is:
AgBr(s) + 2 Na2S2O3(aq) → Na3Ag(S2O3)2(aq) + NaBr(aq)

From the balanced chemical equation, we can see that the molar ratio between AgBr and Na3Ag(S2O3)2 is 1:1. This means that for every 1 mole of AgBr, we will get 1 mole of Na3Ag(S2O3)2.

To calculate the number of moles of AgBr, we need to use its molar mass.
The molar mass of AgBr is:
AgBr = Ag (1 mol) + Br (1 mol) = 107.87 g/mol + 79.90 g/mol = 187.77 g/mol

Now we can calculate the number of moles of AgBr:
Moles of AgBr = Mass of AgBr / Molar mass of AgBr
Moles of AgBr = 0.275 g / 187.77 g/mol ≈ 0.00146 mol

Since the molar ratio between AgBr and Na3Ag(S2O3)2 is 1:1, the number of moles of Na3Ag(S2O3)2 produced will also be 0.00146 mol.

To calculate the mass of Na3Ag(S2O3)2 produced, we need to use its molar mass.
The molar mass of Na3Ag(S2O3)2 is:
Na (3 mol) + Ag (1 mol) + S (2 mol) + O (9 mol)
= 3(22.99 g/mol) + 1(107.87 g/mol) + 2(32.07 g/mol) + 9(16.00 g/mol)
= 68.97 g/mol + 107.87 g/mol + 64.14 g/mol + 144.00 g/mol
= 384.98 g/mol

Now we can calculate the mass of Na3Ag(S2O3)2 produced:
Mass of Na3Ag(S2O3)2 = Moles of Na3Ag(S2O3)2 * Molar mass of Na3Ag(S2O3)2
Mass of Na3Ag(S2O3)2 = 0.00146 mol * 384.98 g/mol ≈ 0.562 g

Therefore, the mass of Na3Ag(S2O3)2 produced if 0.275 g AgBr is removed is approximately 0.562 g.

Answer 4

To determine the mass of Na3Ag(S2O3)2 produced, we need to use stoichiometry and the balanced chemical equation provided. The balanced equation is:

AgBr(s) + 2 Na2S2O3(aq) → Na3Ag(S2O3)2(aq) + NaBr(aq)

From the balanced equation, we can see that the ratio of AgBr to Na3Ag(S2O3)2 is 1:1. This means that for every 1 mole of AgBr, we will produce 1 mole of Na3Ag(S2O3)2.

To find the molar mass of Na3Ag(S2O3)2, we need to sum up the atomic masses of all the elements present:

3(23) + 1(108) + 2(32) + 9(16) = 382 g/mol

Now, let's calculate the moles of Na3Ag(S2O3)2 produced:

moles of Na3Ag(S2O3)2 = mass of AgBr / molar mass of Na3Ag(S2O3)2

moles of Na3Ag(S2O3)2 = 0.275 g / 382 g/mol

moles of Na3Ag(S2O3)2 ≈ 0.000719 moles

Finally, we can calculate the mass of Na3Ag(S2O3)2 produced:

mass of Na3Ag(S2O3)2 = moles of Na3Ag(S2O3)2 × molar mass of Na3Ag(S2O3)2

mass of Na3Ag(S2O3)2 = 0.000719 moles × 382 g/mol

mass of Na3Ag(S2O3)2 ≈ 0.275 g

Therefore, approximately 0.275 g of Na3Ag(S2O3)2 is produced when 0.275 g of AgBr is removed.