# Question 05520

Nov 2, 2016

$\text{3.40 g}$

#### Explanation:

Take a look at the balanced chemical equation that describes this reaction

${\text{Cu"_ ((s)) + 2"AgNO"_ (3(aq)) -> "Cu"("NO"_ 3)_ (2(aq)) + color(blue)(2)"Ag}}_{\left(s\right)}$

Notice that the chemical equation provides you with information about how many moles of each chemical species takes part in this reaction.

More specifically, you know that $1$ mole of solid copper will react with $2$ moles of silver nitrate to produce $1$ mole of aqueous copper(II) nitrate and $\textcolor{b l u e}{2}$ moles of solid silver.

So, assuming that you have enough silver nitrate available to allow for all the sample of copper to react, you can expect the reaction to produce twice as many moles of solid silver than you have moles of solid copper taking part in the reaction.

Use the molar mass of copper to calculate how many moles you have in your sample

1.00 color(red)(cancel(color(black)("g"))) * "1 mole Cu"/(63.546color(red)(cancel(color(black)("g")))) = "0.01574 moles Cu"

According to the aforementioned mole ratio, the reaction will produce

0.01574 color(red)(cancel(color(black)("moles Cu"))) * (color(blue)(2)color(white)(a)"moles Ag")/(1color(red)(cancel(color(black)("mole Cu")))) = "0.03148 moles Ag"

Finally, use the molar mass of silver to convert the number of moles to grams

0.03148color(red)(cancel(color(black)("moles Ag"))) * "107.87 g"/(1color(red)(cancel(color(black)("mole Ag")))) = color(green)(bar(ul(|color(white)(a/a)color(black)("3.40 g")color(white)(a/a)|)))#

The answer is rounded to three sig figs.