a) PTHH: `Cu + 2AgNO_3 -> Cu(NO_3)_2 + 2Ag`
b) Gọi `n_{Cu(pư)} = a (mol)`
Theo PT: `n_{Ag} = 2n_{Cu} = 2a (mol)`
Ta có: `m_{tăng} = m_{Ag} - m_{Cu} = 2a.108 - 64a = 13,2 - 6`
`=>` \(a=\dfrac{9}{190}\left(mol\right)\)
`=>` \(m_{Cu\left(pư\right)}=\dfrac{9.64}{190}=\dfrac{288}{95}\left(g\right)\)
