Đáp án C

$\begin{array}{l}+\left\{\begin{array}{l}I=2A\\ tg=3378s\end{array}\right.\iff n_{electrontraoñoåi}=\frac{3378.2}{96500}=\mathrm{0,07}mol;\left\{\begin{array}{l}Catot:n_{Cu}=\mathrm{0,025}\Rightarrow n_{H_2}=\frac{\mathrm{0,07}-\mathrm{0,025.2}}{2}=\mathrm{0,01}mol\\ Anot:\left\{\begin{array}{l}{n}_{C{l}_2}+n_{O_2}=\mathrm{0,035}-\mathrm{0,01}=\mathrm{0,025}\\ 2n_{C{l}_2}+4n_{O_2}=\mathrm{0,07}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{C{l}_2}=\mathrm{0,015}\\ n_{O_2}=\mathrm{0,01}\end{array}\right.\end{array}\right.\\ \Rightarrow tg=t\left(s\right)anotch\ae giaûiphoùngC{l}_2,catotch\ae giaûiphoùngCu\Rightarrow a=b(*)\Rightarrow n_{electrontraoñoåi}=2a=2b.\\ +tg=2t\left(s\right)\Rightarrow n_{electrontraoñoåi}=4a=4b\Rightarrow \left\{\begin{array}{l}catot:n_{H_2}=\frac{4b-2(b+\mathrm{0,025})}{2}=b-\mathrm{0,025}\\ anot:n_{O_2}=\frac{4a-2(a+\mathrm{0,015})}{4}=\mathrm{0,5}a-\mathrm{0,0075}\end{array}\right.\\ \Rightarrow n_{khíôû2t\left(s\right)}=(b-\mathrm{0,025})+(\mathrm{0,5}a-\mathrm{0,0075})+(a+\mathrm{0,015})=\mathrm{2,0625}a(**)\\ +\left\{\begin{array}{l}(*)\\ (**)\end{array}\right.\Rightarrow \left\{\begin{array}{l}a=\mathrm{0,04}\\ b=\mathrm{0,04}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{Cu{\left(N{O}_3\right)}_2}=\mathrm{0,065}\\ n_{NaCl}=2(\mathrm{0,04}+\mathrm{0,015})=\mathrm{0,11}\end{array}\right.\Rightarrow m=\mathrm{18,655}\Rightarrow gaànnhaátvôùi\end{array}$