Gọi $n_{Fe\ pư} = a(mol)$
\(Fe+CuSO_4\text{→}FeSO_4+Cu\)
a a (mol)
$m_{đinh\ sắt\ sau\ pư} = m_{đinh\ sắt} - m_{Fe\ pư} + m_{Cu}$
$\Rightarrow 8,2 = 7,8 - 56a + 64a$
$\Rightarrow a = 0,05(mol)$
$m_{Cu} = 0,05.64 = 3,2(gam) ; m_{Fe\ pư} = 0,05.56 = 2,8(gam)$