Đáp án:
a,
$3Cu+8HNO_3\to 3Cu(NO_3)_2+2NO+4H_2O$
$MgO+2HNO_3\to Mg(NO_3)_2+H_2O$
$n_{HNO_3}=0,5(mol)$
$n_{NO}=\dfrac{0,672}{22,4}=0,03(mol)$
$\Rightarrow n_{Cu}=\dfrac{3}{2}n_{NO}=0,045(mol)$
$\Rightarrow n_{Cu(NO_3)_2}=n_{Cu}=0,045(mol)$
$\Rightarrow n_{Mg(NO_3)_2}=\dfrac{10,68-0,045.188}{148}=0,015(mol)$
$\Rightarrow n_{MgO}=n_{Mg(NO_3)_2}=0,015(mol)$
Vậy: $\%m_{MgO}=\dfrac{0,015.40.100\%}{0,015.40+0,045.64}=17,24\%$
b,
Ta có: $n_{HNO_3\text{pứ}}=\dfrac{8}{3}n_{Cu}+2n_{MgO}=\dfrac{8}{3}.0,045+2.0,015=0,15(mol)$
$\Rightarrow n_{HNO_3\text{dư}}=0,5-0,15=0,35(mol)$
Vậy $C_{M_{HNO_3}}=\dfrac{0,35}{0,5}=0,7M$