# Given the 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? B. What is the mass of NaBr that will be produced from 42.7 g of AgBr?

Here’s what I got.

The most important thing to look out for when dealing with a problem is if the chemical equation given to you is balanced. As it turns out, that is not the case here, i.e. your equation is unbalanced.

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Given the 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? B. What is the mass of NaBr that will be produced from 42.7 g of AgBr?

##”Na”_ 2″S”_ 2″O”_ (3(aq)) + “AgBr”_ ((s)) -> “NaBr”_ ((aq)) + “Na”_ 3[“Ag”(“S”_ 2″O”_ 3)_ 2] “”_ ((aq))##

In order to balance this equation out, you need ##4## atoms of sodium, ##”Na”##, on the reactants’ side. You also need ##2## thiosulfate ions, ##”S”_2″O”_3^(2-)##, on the reactants’ side. You can thus add a ##color(red)(2)## coefficient in front of sodium thiosulfate, ##”Na”_2″S”_2″O”_3##

##color(red)(2)”Na”_ 2″S”_ 2″O”_ (3(aq)) + “AgBr”_ ((s)) -> “NaBr”_ ((aq)) + “Na”_ 3[“Ag”(“S”_ 2″O”_ 3)_ 2] “”_ ((aq))##

Now, the balanced chemical equation tells you that every mole of silver bromide, ##”AgBr”##, that takes part in the reaction consumes ##color(red)(2)## moles of sodium thiosulfate and produces ##1## mole of sodium bromide, ##”NaBr”##.

The problem gives you grams of silver bromide, so right from the start you know that you must convert them to moles by using the compound’s molar mass

##42.7 color(red)(cancel(color(black)(“g”))) * “1 mole AgBr”/(187.77color(red)(cancel(color(black)(“g”)))) = “0.2274 moles AgBr”##

This many moles of silver bromide will consume

##0.2274 color(red)(cancel(color(black)(“moles AgBr”))) * (color(red)(2)color(white)(a)”moles Na”_2″S”_2″O”_3)/(1color(red)(cancel(color(black)(“mole AgBr”)))) = “0.4548 moles Na”_2″S”_2″O”_3##

and produce

##0.2274color(red)(cancel(color(black)(“moles AgBr”))) * “1 mole NaBr”/(1color(red)(cancel(color(black)(“mole AgBr”)))) = “0.2274 moles NaBr”##

Use the molar mass of sodium bromide to convert the number of moles to grams

##0.2274 color(red)(cancel(color(black)(“moles NaBr”))) * “102.9 g”/(1color(red)(cancel(color(black)(“mole NaBr”)))) = color(green)(|bar(ul(color(white)(a/a)color(black)(“23.4 g”)color(white)(a/a)|)))##

You can thus say that the reaction of ##”42.7 g”## of silver bromide produced ##”23.4 g”## of aqueous sodium bromide and consumed

##”no. of moles of Na”_2″S”_2″O”_3 = color(green)(|bar(ul(color(white)(a/a)color(black)(“0.455 moles”)color(white)(a/a)|)))##

of sodium thiosulfate. Both answers are rounded to three , the number of sig figs you have for the mass of silver bromide.

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