An amount of 1.250g of an alloy (90% Ag, 10% Cu) was treated with HNO3 and H2SO4 to form Ag2SO4 and CuSO4. The solution was then treated with Cu wire to precipitate silver. The resulting solution was treated with excess zinc to precipitate copper. (show formula)

a). What mass of copper will be precipitated?
b). what total mass of zinc will be dissolved (neglect any dissolved by excess H2SO4)? Avoid any unnecessary steps in solving this problem.

thank you!!

To solve this problem, we will follow these steps:

Step 1: Calculate the amount of silver (Ag) and copper (Cu) in the original alloy.
Step 2: Find the amount of Ag2SO4 and CuSO4 formed after treating with HNO3 and H2SO4.
Step 3: Calculate the amount of silver (Ag) precipitated when treated with Cu wire.
Step 4: Determine the amount of copper (Cu) precipitated when treated with excess zinc.
Step 5: Calculate the total mass of zinc (Zn) dissolved.

Let's start with step 1.

Step 1: Calculate the amount of silver (Ag) and copper (Cu) in the original alloy:
Given:
- Total mass of the alloy = 1.250g
- Alloy composition: 90% Ag and 10% Cu

The amount of silver (Ag) in the alloy can be calculated as:
Mass of Ag = (90/100) * Total mass of alloy = (90/100) * 1.250g

The amount of copper (Cu) in the alloy can be calculated as:
Mass of Cu = (10/100) * Total mass of alloy = (10/100) * 1.250g

Now, let's move to step 2.

Step 2: Find the amount of Ag2SO4 and CuSO4 formed after treating with HNO3 and H2SO4:
We are given that the alloy was treated with HNO3 and H2SO4 to form Ag2SO4 and CuSO4. However, we don't need to calculate the exact amount of Ag2SO4 and CuSO4 formed for this particular problem. The information is useful for understanding the reaction that takes place but is not directly required for answering the questions.

Now, let's move to step 3.

Step 3: Calculate the amount of silver (Ag) precipitated when treated with Cu wire:
Since silver (Ag) has a higher reduction potential than copper (Cu), when treated with Cu wire, silver will displace copper from the solution and get precipitated.

Given that the reaction is:
Ag+ + Cu → Cu+ + Ag

The amount of silver (Ag) that will get precipitated is equal to the amount of copper (Cu) present in the original alloy.

Now, let's move to step 4.

Step 4: Determine the amount of copper (Cu) precipitated when treated with excess zinc:
Since copper (Cu) has a lower reduction potential than zinc (Zn), when treated with excess zinc, zinc will displace copper from the solution and get precipitated as CuZn.

The net ionic equation for this reaction is:
Cu2+ + Zn → Zn2+ + Cu

Given that the reaction occurs with excess zinc, all the copper present in the solution will precipitate.

Now, let's move to step 5.

Step 5: Calculate the total mass of zinc (Zn) dissolved:
Since all the copper (Cu) is precipitated, the total mass of zinc (Zn) dissolved is equal to the mass of copper (Cu) present in the original alloy.

Now we can answer the specific questions:

a) What mass of copper will be precipitated?
The mass of copper (Cu) precipitated is equal to the amount of copper present in the original alloy, which we calculated in step 1.

b) What total mass of zinc will be dissolved?
The total mass of zinc (Zn) dissolved is also equal to the amount of copper present in the original alloy, which we calculated in step 1.

In summary:
a) The mass of copper (Cu) precipitated is (10/100) * 1.250g
b) The total mass of zinc (Zn) dissolved is also (10/100) * 1.250g (assuming excess zinc is used)

If you have the specific value for the total mass of the alloy (1.250g in this case), you can substitute it into the formulas to calculate the mass of copper precipitated and the total mass of zinc dissolved.

To solve this problem, we will calculate the mass of copper precipitated and the total mass of zinc dissolved by following the steps described.

Step 1: Calculate the amount of silver in the alloy.
Given that the alloy is 90% Ag, the mass of silver in the alloy is calculated as follows:
Mass of Ag = 90% of 1.250g = 0.9 * 1.250g = 1.125g

Step 2: Calculate the moles of Ag and Cu in the alloy.
The molar mass of Ag = 107.87 g/mol
The molar mass of Cu = 63.55 g/mol

Moles of Ag = Mass of Ag / Molar mass of Ag = 1.125g / 107.87 g/mol = 0.010427 mol
Moles of Cu = Mass of Cu / Molar mass of Cu
Since we have 10% Cu in the alloy, the mass of Cu can be calculated as follows:
Mass of Cu = 10% of 1.250g = 0.1 * 1.250g = 0.125g
Moles of Cu = 0.125g / 63.55 g/mol = 0.001966 mol

Step 3: Determine the stoichiometry of the reactions.
The balanced chemical equations for the reactions are as follows:
For the reaction between Ag in the solution and Cu wire:
2 Ag+ + Cu → Cu2+ + 2Ag
For the reaction between Cu in the solution and excess Zn:
Cu2+ + Zn → Zn2+ + Cu

Step 4: Calculate the moles of silver and copper precipitated.
From the balanced chemical equation, we can see that 2 moles of Ag are formed for every 1 mole of Cu. Therefore, the moles of Ag precipitated will be equal to 2 times the moles of Cu in the alloy:
Moles of Ag precipitated = 2 * Moles of Cu = 2 * 0.001966 mol = 0.003932 mol

Step 5: Calculate the mass of copper precipitated.
Molar mass of Cu = 63.55 g/mol
Mass of Cu precipitated = Moles of Cu precipitated * Molar mass of Cu
Mass of Cu precipitated = 0.001966 mol * 63.55 g/mol = 0.125g

Therefore, the mass of copper precipitated is 0.125g.

Step 6: Calculate the moles of Zn required to dissolve the copper.
From the balanced chemical equation, we can see that 1 mole of Cu is dissolved by 1 mole of Zn. Therefore, the moles of Zn required to dissolve the copper will be equal to the moles of Cu precipitated:
Moles of Zn required = Moles of Cu precipitated = 0.001966 mol

Step 7: Calculate the mass of Zn dissolved.
Molar mass of Zn = 65.38 g/mol
Mass of Zn dissolved = Moles of Zn dissolved * Molar mass of Zn
Mass of Zn dissolved = 0.001966 mol * 65.38 g/mol = 0.1283 g

Therefore, the total mass of zinc dissolved is 0.1283 g.

a). The mass of copper precipitated is 0.125g.
b). The total mass of zinc dissolved is 0.1283g.

1.250g x 0.90 = ??g Ag

1.250 x 0.10 = ??g Cu
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Ag| + HNO3 ==> Ag2SO4
Cu| + HNO3 ==> CuSO4

2Ag^+ + Cu(s)(wire) = Ag(s) + Cu^+2(aq)
Then Zn(s) + Cu^+2(aq) ==> Zn^+2(aq) +
Cu(s)

You know grams Ag initially. Convert to moles and use stoichiometry to calculate the amount of Cu(II) added to the solution when Ag was pptd. Here is a stoichiometry problem if you need it to get started.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Then Cu initially + Cu added to ppt Ag, gives moles Cu reacting with Zn.
Stoichiometry will give you the moles of Zn which can be converted to grams Zn.
In addition to neglecting H2SO4 we must also assume there is no excess of HNO3.
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