How many moles of silver(I) bromide are produced when #6.45# moles of sodium bromide take part in the reaction?
#"NaBr"_ ((aq)) + "AgNO"_ (3(aq)) -> "AgBr"_ ((s)) darr + "NaNO"_ (3(aq))#
1 Answer
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
#"NaBr"_ ((aq)) + "AgNO"_ (3(aq)) -> "AgBr"_ ((s)) darr + "NaNO"_ (3(aq))#
Stoichiometric coefficients of
#1"NaBr"_ ((aq)) + 1"AgNO"_ (3(aq)) -> 1"AgBr"_ ((s)) darr + 1"NaNO"_ (3(aq))#
So, this tells you that for every mole of sodium bromide,
Since you know that
#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)|)))#