\\\\(\\underset {\\underset {= 28.0\\ g}{1\\ mol}}{Na(g)} + \\underset {\\underset {= 6.0\\ g}{3\\ mol}}{3H_{2(g)}}\\rightarrow \\underset {\\underset {= 34.0\\ g}{2\\ mol}}{2NH_{3(g)}}\\\\)(i) \\\\(28.0\\ g\\\\) of \\\\(N_{2}\\\\) require \\\\(6.0\\ g\\\\) of \\\\(H_{2}\\\\) to produce\\\\(= 34.0\\ g\\\\) of \\\\(NH_{3}\\\\)\\\\(2.00\ imes 10^{3}g\\\\) of \\\\( N_{2}\\\\) will produce\\\\(= \\dfrac {34}{28} \ imes 2.00\ imes 10^{3}g\\\\) of \\\\(NH_{3}\\\\)\\\\(= 2.43\ imes 10^{3}g\\\\) of \\\\(NH_{3}\\\\)\\\\(= 2430\\ g\\ NH_{3}\\\\)(ii) Yes, dihydrogen will remain unreacted to some extent(iii) Amount of hydrogen that remains unreacted.\\\\(28.0\\ g\\\\) of \\\\(N_{2}\\\\) require \\\\(6.0\\ g\\\\) of \\\\(H_{2}\\\\)\\\\(2.00\\ g\ imes 10^{3}g\\\\) of \\\\(N_{2}\\\\) will require\\\\(= \\dfrac {6.0}{28.0}\ imes 2.00 \ imes 10^{3}\\\\) of \\\\(H_{2} = 428.5\\ g\\\\) of \\\\(H_{2}\\\\)\\\\(\ herefore\\\\) Amount of hydrogen that remains unreacted\\\\(= [1.00\ imes 10^{3} - 428.5]_{g}\\\\)\\\\(= 571.5\\ g\\\\).