# Question 93ac1

Jul 14, 2016

Here's what I got.

#### Explanation:

The most important thing to look out for when dealing with a stoichiometry 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.

${\text{Na"_ 2"S"_ 2"O"_ (3(aq)) + "AgBr"_ ((s)) -> "NaBr"_ ((aq)) + "Na"_ 3["Ag"("S"_ 2"O"_ 3)_ 2] }}_{\left(a q\right)}$

In order to balance this equation out, you need $4$ atoms of sodium, $\text{Na}$, on the reactants' side. You also need $2$ thiosulfate ions, ${\text{S"_2"O}}_{3}^{2 -}$, on the reactants' side. You can thus add a $\textcolor{red}{2}$ coefficient in front of sodium thiosulfate, ${\text{Na"_2"S"_2"O}}_{3}$

$\textcolor{red}{2} {\text{Na"_ 2"S"_ 2"O"_ (3(aq)) + "AgBr"_ ((s)) -> "NaBr"_ ((aq)) + "Na"_ 3["Ag"("S"_ 2"O"_ 3)_ 2] }}_{\left(a q\right)}$

Now, the balanced chemical equation tells you that every mole of silver bromide, $\text{AgBr}$, that takes part in the reaction consumes $\textcolor{red}{2}$ moles of sodium thiosulfate and produces $1$ mole of sodium bromide, $\text{NaBr}$.

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 $\text{42.7 g}$ of silver bromide produced $\text{23.4 g}$ 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 sig figs, the number of sig figs you have for the mass of silver bromide.