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stiks02 [169]
3 years ago
13

Copper metal (Cu) reacts with silver nitrate (AgNO3) in aqueous solution to form Ag and Cu(NO3)2. An excess of AgNO3 is present.

The balanced chemical equation is shown below.
Cu + 2AgNO3 ---> Cu(NO3)2 + 2Ag

The molar mass of Cu is 63.5 g/mol. The molar mass of Ag is 107.9 g/mol. What mass, in grams, of Ag is produced from reaction of 31.75 g of Cu?

26.95
107.9
215.91
431.82
Chemistry
2 answers:
aev [14]3 years ago
8 0


we are given the reaction Cu + 2AgNO3 ---> Cu(NO3)2 + 2Ag. This means for every mole Cu used, there are 2 moles of Ag produced. In this case, given 31.75 g Cu, converting to moles through molar mass and using stoichiometric ratio and the molar mass of Ag, the mass Ag produced is 107.9 grams. 
Flauer [41]3 years ago
7 0

The mass in grams of Ag produced from reaction of 31.75g of Cu is 107.9 grams

calculation

Cu+2AgNO3 ---> Cu(NO3)2 + 2Ag

find the moles of Cu that reacted

moles= mass/molar mass

=31.75 g/63.5 g/mol= 0.5 moles

from the equation above the mole ratio of CU:Ag is 1:2 therefore the moles of Ag=0.5 moles x2 = 1 mole

mass of Ag= moles of Ag x molar mass

= 1 mole x107.9g/mol = 107.9 grams

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lilavasa [31]

1.66 M is the concentration of the chemist's working solution.

<h3>What is molarity?</h3>

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.

In this case, we have a solution of Zn(NO₃)₂.

The chemist wants to prepare a dilute solution of this reactant.

The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.

We want to know the concentration of this diluted solution.

As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:

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n = M x V_2

If we replace this expression in (1) we have:

M_1 x V_1= M_2 x V_2

Where 1, would be the stock solution and 2, the solution we want to prepare.

So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is M_2:

4.93 x 210 =  620 xM_2

M_2 = 1.66 M

This is the concentration of the solution prepared.

Learn more about molarity here:

brainly.com/question/19517011

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