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

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

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Answer 1 · 2015-05-15T00:55:22

You could make 863.0 g of silver.

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

$C u_{(s)} + 2 A g N O_{3 (a q)} \to C u {(N O_{3})}_{2 (a q)} + 2 A g_{(s]}$ $Cu_((s)) + color(red)(2)AgNO_(3(aq)) -> Cu(NO_3)_(2(aq)) + color(blue)(2)Ag_text((s])$

Notice the $1 : 2$ $1:color(red)(2)$ 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

$381 g \cdot \frac{1 mole Cu}{63.55 g} = 6.00 moles Cu$ $381cancel("g") * "1 mole Cu"/(63.55cancel("g")) = "6.00 moles Cu"$

$1359 g \cdot \frac{1 mole A g N O_{3}}{169.87 g} = 8.000 moles$ $1359cancel("g") * ("1 mole "AgNO_3)/(169.87cancel("g")) = "8.000 moles"$ $A g N O_{3}$ $AgNO_3$

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

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

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

Notice that you have a $1 : 1$ $1:1$ mole ratio ( $2 : 2$ $color(red)(2):color(blue)(2)$ ) 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

$8.000 moles A g N O_{3} \cdot \frac{1 mole Ag}{1 mole A g N O_{3}} = 8.000 moles Ag$ $8.000cancel("moles "AgNO_3) * ("1 mole Ag")/(1cancel("mole "AgNO_3)) = "8.000 moles Ag"$

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

$8.000 moles Ag \cdot \frac{107.87 g}{1 mole Ag} = 862.96 g$ $8.000cancel("moles Ag") * "107.87 g"/(1cancel("mole Ag")) = "862.96 g"$

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

$m_{A g} = 863.0 g$ $m_(Ag) = color(green)("863.0 g")$

Answer 2 · 2015-05-15T01:03:10

864g

$C u_{(s)} + 2 A g N O_{3 (a q)} \to C u {(N O_{3})}_{2 (a q)} + 2 A g ⏐ ⏐ ↓$ $Cu_((s))+2AgNO_(3(aq))rarrCu(NO_3)_(2(aq))+2Agdarr$

$1 mol + 2 mol \to 2 mol$ $1 "mol" + 2 "mol"rarr2"mol"$

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 $\to$ $rarr$ 8 mol Ag

8 mol Ag = 108 x 8 = 864g

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

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