Answered

If 9.85 grams of copper metal react with 31.0 grams of silver nitrate, how many grams of copper nitrate can be formed and how many grams of the excess reactant will be left over when the reaction is complete? Show all of your work.


unbalanced equation: Cu + AgNO3 yields Cu(NO3)2 + A

Verified Answer :

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The balanced chemical reaction:

Cu + 2AgNO3 = Cu(NO3)2 + 2Ag

We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.

9.85 g Cu ( 1 mol Cu / 63.55 g Cu ) = 0.15 mol Cu
31.0 g AgNO3 ( 1 mol AgNO3 / 169.87 g AgNO3 ) = 0.18 mol AgNO3

The limiting reactant is AgNO3.

0.18 mol AgNO3 ( 1 mol Cu(NO3)2 / 2 mol AgNO3 ) (187.56 g / 1 mol) =16.88 g Cu(NO3)2

0.15 mol Cu - 0.18 mol AgNO3 ( 1 mol Cu / 2 mol AgNo3) = 0.06 mol Cu excess

0.06 mol Cu ( 63.55 g Cu / 1 mol Cu ) = 3.81 g Cu excess

Answer : The mass of excess reactant will be left over is 4.03 g and mass of \(Cu(NO_3)_2\)  is 34.2 g.

Solution : Given,

Mass of Cu = 9.85 g

Mass of \(AgNO_3\) = 31.0 g

Molar mass of Cu = 64 g/mole

Molar mass of \(AgNO_3\) = 170 g/mole

Molar mass of \(Cu(NO_3)_2\) = 188 g/mole

First we have to calculate the moles of Cu and \(AgNO_3\).

\(\text{ Moles of }Cu=\frac{\text{ Mass of }Cu}{\text{ Molar mass of }Cu}=\frac{9.85g}{64g/mole}=0.154moles\)

\(\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{31.0g}{170g/mole}=0.182moles\)

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

\(Cu+2AgNO_3\rightarrow Cu(NO_3)_2+2Ag\)

From the balanced reaction we conclude that

As, 2 mole of \(AgNO_3\) react with 1 mole of \(Cu\)

So, 0.182 moles of \(AgNO_3\) react with \(\frac{0.182}{2}=0.091\) moles of \(Cu\)

From this we conclude that, \(Cu\) is an excess reagent because the given moles are greater than the required moles and \(AgNO_3\) is a limiting reagent and it limits the formation of product.

Moles of remaining excess reactant = 0.154 - 0.091 = 0.063 moles

Now we have to calculate the mass of excess reactant will be left over.

\(\text{ Mass of }Cu=\text{ Moles of }Cu\times \text{ Molar mass of }Cu\)

\(\text{ Mass of }Cu=(0.063moles)\times (64g/mole)=4.03g\)

Now we have to calculate the moles of \(Cu(NO_3)_2\)

From the reaction, we conclude that

As, 2 mole of \(AgNO_3\) react to give 1 mole of \(Cu(NO_3)_2\)

So, 0.182 moles of \(AgNO_3\) react to give \(\frac{0.182}{2}=0.091\) moles of \(Cu(NO_3)_2\)

Now we have to calculate the mass of \(Cu(NO_3)_2\)

\(\text{ Mass of }Cu(NO_3)_2=\text{ Moles of }Cu(NO_3)_2\times \text{ Molar mass of }Cu(NO_3)_2\)

\(\text{ Mass of }Cu(NO_3)_2=(0.182moles)\times (188g/mole)=34.2g\)

Therefore, the mass of \(Cu(NO_3)_2\)  is 34.2 grams.