Question

Mole to Mole relationship between Cu and Ag

I answered 1- a) thru c) this is the results from the lab I need help with the questions, please.

185.000 g = empty beaker

390.233 g= beaker with 200 ml of silver nitrate

391.233 g= beaker with 200 ml of silver nitrate + 1.000 ml of copper

188.395 g = beaker with solid substance after decanting silver nitrate and copper

387.388g = beaker with liquid substance after decanting silver nitrate and copper

1. Record and calculate the following (make sure to show your work on (d) & (e)):

a) mass of silver nitrate solution [mass of the silver nitrate solution in the beaker minus the mass of the empty beaker] (g)

390.233-185.000= 205.233g

b) mass of copper solution produced in the reaction [mass of the copper solution in the beaker minus the mass of the empty beaker](g)

391.233-185.000= 206.233

c mass of silver produced by the reaction [mass of the solid silver in the beaker minus the mass of the empty beaker] (g)

188.395-185.000= 3.395

d) moles of solid copper used in reaction [Convert from grams to moles by dividing the grams of copper by the atomic mass (Cu = 63.55 g/mol)] (mol)

e) moles of solid silver produced in reaction [Convert from grams to moles by dividing the grams of silver by the atomic mass (Ag = 107.84 g/mol)] (mol)

2. Write the equation for the reaction between copper and silver ion. Include your experimentally determined mole ratios as fractional coefficients. (The equation without the coefficients is in the background. To write the equation with fractional coefficients, simply place the moles you calculated for Cu (d) in front of copper and copper ion and do the same thing for silver and silver ion (e)).

3. Convert the fractional coefficients to a whole number ratio and rewrite the equation using the whole number ratio. (To convert fractional coefficients to a whole number ratio, you will need to divide each reactant and product with the smallest moles. For example, I have an equation: 0.1 Zn (s) + 0.5 Ag + (aq)  0.1 Zn +2 (aq) + 0.5 Ag(s). In this case the smallest mole is 0.1. Divide each moles with 0.1, then the whole number ratio would be Zn (s) + 5Ag + (aq)  Zn +2 (aq) + 5Ag(s))

4. If the silver in the beaker contained water during your last weighing, how would this affect your results?

5. Assume that magnesium would act atom-for-atom exactly the same as copper in this experiment. How many grams of magnesium would, have been used in the reaction if one gram of silver were produced? The atomic mass of magnesium is 24.31 g/mol. (Show work here)(Hints: You have to write a balanced equation to calculate this. To write the equation, replace copper and copper ion with magnesium and magnesium ion.)

6. Explain the source of the blue color of the solution after the reaction.

Answer 1

a) Mass of silver nitrate solution [mass of the silver nitrate solution in the beaker minus the mass of the empty beaker] (g)

390.233-185.000= 205.233g

b) Mass of copper solution produced in the reaction =

([mass of the copper solution in the beaker minus the mass of the empty beaker](g)

391.233-185.000= 206.233 is not correct)

Correct answer is

Mass of copper solution produced in the reaction =[ beaker with 200 ml of silver nitrate + 1.000 ml of copper - beaker with 200 ml of silver nitrate] = [391.233-390.233] = 1.0 g

c Mass of silver produced by the reaction [mass of the solid silver in the beaker minus the mass of the empty beaker] (g)

188.395-185.000= 3.395

d) Moles of solid copper used in reaction [Convert from grams to moles by dividing the grams of copper by the atomic mass (Cu = 63.55 g/mol)] (mol)

Moles of cu = given mass of cu/ molar mass

       = 1.0/63.55 g/mol = 0.0157 moles

e)

Moles of solid silver produced = given mass of silver/molar mass

                                                       = 3.395/107.84 = 0.0315 mol

2) Write the equation for the reaction between copper and silver ion. Include your experimentally determined mole ratios as fractional coefficients.

0.0157 Cu (s) + 0.0315 Ag⁺à 0.0157 Cu⁺² + 0.0315 Ag (s)

3) Convert the fractional coefficients to a whole number ratio and rewrite the equation using the whole number ratio.

Moles of copper are smaller than moles of silver, so divide moles of silver by moles of copper to convert the fractional coefficients to a whole number ratio.

0.0315 Ag+/0.0157 Cu (s) = 2.0

The equation using the whole number ratio Cu(s) +2Ag ⁺ (aq) --> Cu ²⁺ (aq) + 2Ag(s)

4. If the silver in the beaker contained water during your last weighing, the mass of silver would end up weighing more than it should. Moles of silver would also be more, affecting the molar ratios needed in order to calculate the balanced equation.

5. Write a balanced equation

Mg(s) +2Ag + (aq) --> Mg²⁺ (aq) + 2Ag(s)

The atomic mass of magnesium is 24.31 g/mol and the atomic mass of silver is 107.84g/mol

1 g Ag of silver is produced from [(1 mol Mg)(24.31 g/mol)] /[ (2mol Ag)(107.84g/mol)]

1 g Ag of silver is produced from 0.1127 grams of Mg

0.1127 grams of Mg is used to produce one gram of silver

6. The blue color of the solution is due to presence of Cu²⁺ and water. Cu²⁺ in water form hydrate, which is blue in color.