Question

8. What is the percent yield if 52.36 grams of Na2S2O3 reacts with 29.47 grams of AgBr to form 6.15 g of NaBr? s vodo The following information may be helpful with this problem 9. Molar mass of AgBr = 187.78 g/mol. Molar mass of K2S2O3 = 190.31 g/mol. Molar mass of KBr = 126.00 g/mol. Molar mass of K3Ag(S203)2 = 470.40 g/mol. 9. Consider the following balanced equation: 2K2S2O3 + AgBr K3Ag(S203)2 + KBr How many grams of K2S2O3 are needed to react with 8.34 grams of AgBr?

Answer 1

Answer 8-

Given,

Mass of Na₂S₂O₃ reacted = 52.36 g

Mass of AgBr reacted = 29.47 g

Mass of NaBr Formed (Actual) = 6.15 g

Molar Mass of AgBr = 187.77 g/mol

Molar Mass of NaBr = 102.894 g/mol

Molar Mass of Na₂S₂O₃ = 158.11 g/mol

% Yield = ?

AgBr + 2 Na₂S₂O₃  $\rightarrow$ Na₃[Ag(S₂O₃)₂] + NaBr

We know that,

Moles = Mass/ Molar Mass

Moles of AgBr = 29.47 g/187.77 g/mol = 0.157 mol

Moles of Na₂S₂O₃ = 52.36 g/158.11 g/mol = 0.331 mol-----A

AgBr + 2 Na₂S₂O₃  $\rightarrow$ Na₃[Ag(S₂O₃)₂] + NaBr

For limiting reagent,

Divide the stiochiometric coefficient by the moles. The one with smaller ratio is the LIMITING REAGENT.

For AgBr = 1/0.157 = 6.37

For Na₂S₂O₃ = 6.04

So, Na₂S₂O₃ = LIMITING REAGENT

Also,

Using Stiochiometry, it can be analyzed that for 2 mole of Na₂S₂O₃, 1 mole of NaBr is produced.

Moles of NaBr produced = (1/2) * Moles Na₂S₂O₃

So,

Moles of NaBr produced = (1/2)*0.331 mol

Moles of NaBr produced = 0.1655 mol

Now,

Moles = Mass/ Molar Mass

Mass = Moles * Molar Mass

So , Mass of NaBr in 0.1655 mol = 0.1655 mol * 102.894 g/mol

Mass of NaBr in 0.1655 mol (Theoretical) = 17.02 g

Now,

Percentage Yield = (Actual Yield/ Theorectical Yield)*100

Percentage Yield = (6.15 g/ 17.02 g)*100

Percentage Yield = 36.13 % [Answer]

Answer 9 -

Given,

Mass of AgBr = 8.34 g

Molar Mass of AgBr =  187.78 g/mol

Molar Mass of K₂S₂O₃ = 190.31 g/mol

Mass of K₂S₂O₃ needed = ?

AgBr + 2 K₂S₂O₃  $\rightarrow$ K₃[Ag(S₂O₃)₂] + KBr

Using Stiochiometry it can be analyzed that for 1 mole of AgBr, 2 moles of K₂S₂O₃ are required

So, moles of K₂S₂O₃ are required = 2 * moles of AgBr

Now,

Moles = Mass/ Molar Mass

Moles of AgBr = 8.34 g/187.78 g/mol

Moles of AgBr = 0.044 mol

So, moles of K₂S₂O₃ are required = 2 * 0.044 mol = 0.089 mol

Also,

Mass = Moles * Molar Mass

Mass of of K₂S₂O₃ are required = 0.089 mol * 190.31 g/mol

Mass of of K₂S₂O₃ are required = 16.93 g [Answer]