Answer to Question #65656 in General Chemistry for Briana

Question

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

Copper reacts with silver nitrate to form silver and copper(II) nitrate. How many grams of silver would form if 4.80e+19 atoms of copper reacts with excess silver nitrate?

What mass of silver nitrate is needed to react completely with the copper?

Answer 1

Cu + 2AgNO3 → 2Ag +Cu(NO3)2
N(Cu)=4.80*1019
NA =6.02*1023 mol−1
n(Cu)=N/N_A =〖4.80*10〗^19/〖6.02*10〗^23 =0.8*10-4 mol
n(Ag)=2n(Cu)=2*0.8*10-4=1.6*10-4 mol
n(Ag)=(m(Ag))/(M(Ag))
M(Ag)=108 g/mol
m(Ag)=n(Ag)*M(Ag)=1.6*10-4*108=0.0173 g
M(AgNO3)=M(Ag)+M(N)+3M(O)=108+14+3*16=170 g/mol
m(AgNO3)=n(AgNO3)*M(AgNO3)=1.6*10-4*170=0.0272 g
m(Ag)= 0.0173 g – mass of silver
m(AgNO3)= 0.0272 g mass of silver nitrate