How many moles of silver(I) bromide are produced when 6.45 moles of sodium bromide take part in the reaction?

${\text{NaBr"_ ((aq)) + "AgNO"_ (3(aq)) -> "AgBr"_ ((s)) darr + "NaNO}}_{3 \left(a q\right)}$

Jun 21, 2016

$\text{6.45 moles AgBr}$

Explanation:

The thing to remember about balanced chemical equations is that they tell you the mole ratios in which the reactants and the products find themselves for a given chemical equation.

More specifically, the balanced chemical equation tells you how many moles of each reactant will react with each other and how many moles of products will be produced as a result.

In this case, the balanced chemical equation that describes this double replacement reaction looks like this

${\text{NaBr"_ ((aq)) + "AgNO"_ (3(aq)) -> "AgBr"_ ((s)) darr + "NaNO}}_{3 \left(a q\right)}$

Stoichiometric coefficients of $1$ are omitted from the balanced chemical equation, but in your case the problem wants to give you a hand and lists them

$1 {\text{NaBr"_ ((aq)) + 1"AgNO"_ (3(aq)) -> 1"AgBr"_ ((s)) darr + 1"NaNO}}_{3 \left(a q\right)}$

So, this tells you that for every mole of sodium bromide, $\text{NaBr}$, that takes part in the reaction you get $1$ mole of silver bromide, $\text{AgBr}$ $\to$ the two compounds are in a $1 : 1$ mole ratio.

Since you know that $6.45$ moles of sodium bromide are available for the reaction, and assuming that all of them react, you can say that the reaction will produce

6.45 color(red)(cancel(color(black)("moles NaBr"))) * "1 mole AgBr"/(1color(red)(cancel(color(black)("mole NaBr")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("6.45 moles AgBr")color(white)(a/a)|)))