a,
Gọi x, y là số mol Fe, Cu.
$\Rightarrow 56x+64y=15$ (1)
$Fe+CuSO_4\to FeSO_4+Cu$
Sau phản ứng: $n_{Cu}=x+y (mol)$
$\Rightarrow 64x+64y=16,5$ (2)
$(1)(2)\Rightarrow x=0,1875; y=0,0703125$
$\%m_{Fe}=\dfrac{0,1875.56.100}{15}=70\%$
$\%m_{Cu}=30\%$
b,
$n_{Cu}=x+y=0,2878125(mol)$
$3Cu+8HNO_3\to 3Cu(NO_3)_2+2NO+4H_2O$
$\Rightarrow n_{HNO_3}=\dfrac{8}{3}n_{Cu}=0,6875(mol)$
$\Rightarrow V_{HNO_3}=\dfrac{0,6875+0,6875.10\%}{2}=0,378125l$