Đáp án đúng: C 0,1 $\underset{0,2\,mol}{\mathop{4Mg}}\,\,+\,10HN{{O}_{3}}\,\xrightarrow{{}}\,\underset{\xrightarrow{{}}}{\mathop{4Mg{{(N{{O}_{3}})}_{2}}}}\,\,+\,\underset{0,05\,\,mol}{\mathop{{{N}_{2}}O}}\,\,+\,5{{H}_{2}}O$ ${{n}_{HN{{O}_{3}}}}$ bị khử =$2{{n}_{{{N}_{2}}O}}$= 0,1 mol