Na2S2O3+AgBr=NaBr+Na3[Ag(S2O3)]2 How many moles of Na2S2O3 are needed to react completely with 42.7g of AgBr? What is the mass of NaBr that will be produced from 42.7 g of AgBr?

Question

Step by step and where each number came from and clear explanation

Answer 1

The reaction as given is

Na₂S₂O₃ + AgBr ==> NaBr + Na₃[Ag(S₂O₃)₂] (note you wrote it incorrectly in the question - be careful)

Step 1: Balance the equation

2Na₂S₂O₃ + AgBr ==> Na₃[Ag(S₂O₃)₂] + NaBr

Step 2: Calculate moles of AgBr present

42.7 g AgBr x 1 mole AgBr/187.8 g = 0.227 moles AgBr

Step 3: Use the mole ratio of the balanced equation to calculate moles Na₂S₂O₃ needed

The mole ratio of Na₂S₂O₃ to AgBr is 2:1

0.227 moles AgBr x 2 moles Na₂S₂O₃ / mole AgBr = 0.454 moles Na₂S₂O₃ needed (answer to part 1)

Step 4: Use the mole ratio and molar mass of NaBr to calculate mass of NaBr produced

0.227 moles AgBr x 1 mol NaBr/mol AgBr x 103 g NaBr/mol NaBr = 23.4 g NaBr produced (answer to part 2)