5. 1.250 g of an alloy which is 90.00 % Ag and 10.00 % Cu is reacted with nitric acid producing AgNO3 and Cu(NO3)2 dissolved in solution. To the solution of the nitrate compounds is added a piece of copper wire that reacts only with the AgNO3 in solution to precipitate all of the silver, and produce additional Cu(NO3)2 (aq) . (The equation for this single-replacement reaction is: AgNO3(aq) + Cu(s)  Cu(NO3)2(aq) + 2 Ag(s) ) The resulting solution containing all of the Cu(NO3)2 produced by the above processes is treated with excess zinc to precipitate the entire amount of copper (ll) ion present. ( The equation for this step is Cu(NO3)2(aq) + Zn(s)  Zn(NO3)2(aq) + Cu(s) ) What mass of copper will be precipitated?

See explanation

Explanation:

According to the question Copper is obtained from  the original alloy and the copper wire used to displace the silver.  

So,

Mass of the copper from the original alloy: (10.00 % * 1.250 g) =  0.125 g

We now need to obtain the mass of the copper from the wire, then we consider how much silver is precipitated:

Mass of silver in the alloy; 90.00 % * 1.250 g = 1.125 g

So,

Amount of silver precipitated =

1.125 / 107.9 = 0.01043 mole

From the reaction equation, it takes two moles of silver to displace one mole of copper.

Hence, the amount of copper from the wire is:  0.01043 / 2 = 0.005213 mol

And a mass of  0.005213(63.55) = 0.3313 g

Total mass of copper=  0.125 + 0.3313 = 0.456 g

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