a)
$Zn + 4HNO_3 \to Zn(NO_3)_2 + 2NO_2 + 2H_2O$
$Cu + 4HNO_3 \to Cu(NO_3)_2 + 2NO_2 + 2H_2O$
b)
Gọi $n_{Zn} = a(mol) ; n_{Cu} = b(mol)$
Ta có :
$65a + 64b = 3,23$
$n_{NO_2} = 2a + 2b = 0,1$
$\Rightarrow a = 0,03 ; b = 0,02$
$\%m_{Zn} = \dfrac{0,03.65}{3,23}.100\% = 60,37\%$
$\%m_{Cu} = 100\% -60,37\% = 39,63\%$
c)
$n_{HNO_3} = 2n_{NO_2} = 0,2(mol)$
$C_{M_{HNO_3}} = \dfrac{0,2}{0,1} = 2M$
$m_{Zn(NO_3)_2} = 0,03.189 = 5,67(gam)$
$m_{Cu(NO_3)_2} = 0,02.188 = 3,76(gam)$