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QUESTION

Given the reaction: Na2S2O3 + AgBr ----&gt; 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.

"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.