Question

Given the following reaction: Na2S2O3 + AgBr NaBr + Na3[Ag(S2O3)2] a. How many moles of Na2S2O3 are needed to react completely with 42.7 g of AgBr?

Answer 1

0.2274 mole

Step-by-step explanation:

Calculation of the moles of
`AgBr` as:-

Mass = 42.7 g

Molar mass of
`AgBr` = 187.77 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (42.7\ g)/(187.77\ g/mol)`

`Moles= 0.2274\ mol`

According to the reaction shown below:-

`Na_2S_2O_3 + AgBr\rightarrow NaBr + Na_3[Ag(S_2O_3)_2]`

1 mole of
`AgBr` reacts with 1 mole of
`Na_2S_2O_3`

So,

0.2274 mole of
`AgBr` reacts with 0.2274 mole of
`Na_2S_2O_3`

Moles of
`Na_2S_2O_3` needed = 0.2274 mole