# If you have 381 g Cu and 1359 g AgNO3, how many grams of silver could be made?

May 15, 2015

You could make 863.0 g of silver.

Start by writing the balanced chemical equation for this single replacement reaction

$C {u}_{\left(s\right)} + \textcolor{red}{2} A g N {O}_{3 \left(a q\right)} \to C u {\left(N {O}_{3}\right)}_{2 \left(a q\right)} + \textcolor{b l u e}{2} A {g}_{\textrm{\left(s\right]}}$

Notice the $1 : \textcolor{red}{2}$ mole ratio that exists between copper and silver nitrate. This tells you that you need 2 moles of silver nitrate for every mole of copper that takes part in the reaction.

Anytthing less than that, and silver chloride will act as a limiting reagent.

Use the molar masses of copper and of silver nitrate to determine how many moles of each you have

381cancel("g") * "1 mole Cu"/(63.55cancel("g")) = "6.00 moles Cu"

1359cancel("g") * ("1 mole "AgNO_3)/(169.87cancel("g")) = "8.000 moles" $A g N {O}_{3}$

Notice that you have insufficient silver nitrate to allow for all of the moles of copper to react. That much copper would have needed

6.00cancel("moles Cu") * (color(red)(2)"moles "AgNO_3)/(1cancel("mole Cu")) = "12.0 moles" $A g N {O}_{3}$

As a result, all of the silver nitrate will react, and you'll be left with excess copper metal.

Notice that you have a $1 : 1$ mole ratio ($\textcolor{red}{2} : \textcolor{b l u e}{2}$) between silver nitrate and silver metal; this means that the number of moles of silver the reaction will produce will be equal to the number of moles of silver nitrate that react.

As a result, you'll get

8.000cancel("moles "AgNO_3) * ("1 mole Ag")/(1cancel("mole "AgNO_3)) = "8.000 moles Ag"

Now use silver's molar mass to determine how many grams you'd produce

8.000cancel("moles Ag") * "107.87 g"/(1cancel("mole Ag")) = "862.96 g"

Rounded to four sig figs, the number of sig figs given for the mass of silver nitrate, the answer will be

${m}_{A g} = \textcolor{g r e e n}{\text{863.0 g}}$

May 15, 2015

864g

$C {u}_{\left(s\right)} + 2 A g N {O}_{3 \left(a q\right)} \rightarrow C u {\left(N {O}_{3}\right)}_{2 \left(a q\right)} + 2 A g \downarrow$

$1 \text{mol" + 2 "mol"rarr2"mol}$

Now we have 281g Cu = 381/63.5 = 6 mol.

We have 1359g silver nitrate = 1359/170 = 8 mol

But 6 mol Cu requires 12 mol silver nitrate.

So it is the 8 mol of silver nitrate which will decide the yield of the reaction since Cu is in XS.

So 8 mol silver nitrate $\rightarrow$ 8 mol Ag

8 mol Ag = 108 x 8 = 864g