cu+2agno3-- cu(no3)2+2ag how much silver can be made from 17 g of silver nitrate?

'Cu + 2agno3 .. cu(no3)2 + 2ag '
First of all to correct your equation . Elemental symbols of ONE letter us a capital letter.
Elemental symbols of TWO letters use a first letter as capital and second letter as small case.
Hence
'Cu + 2AgNO3 = Cu(NO3)2 + 2Ag

The molar ratios are 1:2::1:2
Next calculate the moles of Silver Nitrate (AgNO3)
Ag x 1 = 108 x 1 = 108
N x 1 = 14 x 1 = 14
O x 3 = 16 x 3 = 48
108 + 14 + 48 = 170
moles(AgNO3) = 17g / 170 = 0.1 moles ( equivalent to '2' in the molar ratios above).
moles (Ag) = 0.1 ( Also , equivalent to '2' in the molar ratios above).

moles(Ag) = 0.1 = mss(g) / 108
Algebraically rearrange
mass(g) = 108 X 0.1 = 10.8 g (Ag produced).

Molar mass of Ag = 108 g/mol
Molar mass of AgNO₃ = 108 + 14 + 16×3 = 170 g/mol

Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag

No. of moles of AgNO₃ reacted = (17 g)/(170 g/mol) = 0.1 mol
No. of moles of Ag formed = (0.1 mol) × (2/2) = 0.1 mol
Mass of Ag formed = (108 g/mol) × (0.1 mol) = 10.8 g

n(AgNO3)=0.1 moles => n(Ag)=0.2 moles

m(Ag)=21.6 g

2agno3 Cu