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ANEK [815]
3 years ago
11

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 ma

ny 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
Chemistry
2 answers:
professor190 [17]3 years ago
7 0

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.

goldfiish [28.3K]3 years ago
3 0
The balanced chemical reaction:

<span>Cu + 2AgNO3 = Cu(NO3)2 + 2Ag
</span>
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

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