Đáp án:
Ta có: $\dfrac{n_{NaCl}}{n_{Cu(NO_3)_2}}=\dfrac{0,5}{2}=\dfrac{1}{4}$
PT điện phân:
$NaCl+H_2O\to NaOH+\dfrac{1}{2}H_2+\dfrac{1}{2}Cl_2$
$4Cu(NO_3)_2+4H_2O\to 4Cu+8HNO_3 + 2O_2$
Cộng hai PT ta có:
$NaCl+4Cu(NO_3)_2+6H_2O\to 4Cu+ NaOH+ 8HNO_3 + \dfrac{1}{2}H_2+\dfrac{1}{2}Cl_2+2O_2$
$\Rightarrow NaCl+4Cu(NO_3)_2+5H_2O \to 4Cu+NaNO_3+7HNO_3+\dfrac{1}{2}H_2+\dfrac{1}{2}Cl_2+2O_2$