Đáp án đúng: D
41,2.
$\displaystyle \underset{x\,mol}{\mathop{{{(N{{H}_{4}})}_{2}}C{{O}_{3}}}}\,\,\xrightarrow{{{t}^{o}}}\,\underset{x\,mol}{\mathop{N{{H}_{3}}}}\,\,+\,\underset{x\,mol}{\mathop{N{{H}_{4}}HC{{O}_{3}}}}\,$
$\underset{(y\,+\,x)\,mol}{\mathop{N{{H}_{4}}HC{{O}_{3}}}}\,\,\xrightarrow{{{t}^{o}}}\,\underset{y\,+\,x}{\mathop{N{{H}_{3}}}}\,\,+\underset{y\,+\,x}{\mathop{C{{O}_{2}}}}\,\,+\,{{H}_{2}}O$
$\Rightarrow \,\left\{ \begin{array}{l}{{n}_{N{{H}_{3}}}}\,=\,x\,+\,y\,+\,x\,=\,0,6\,mol\\{{n}_{C{{O}_{2}}}}\,=\,y\,+\,x\,=\,0,5\,mol\end{array} \right.\,\Rightarrow \,\left\{ \begin{array}{l}x\,=\,0,1\\y\,=\,0,4\end{array} \right.$
$\Rightarrow \,m\,=\,0,1.96\,+\,0,4.79\,=\,41,2\,gam$
