Convert both masses to moles and compare them with the 1:3 stoichiometric ratio. Nitrogen is limiting.
\[\begin{aligned}n(N_2) &= \frac{2000}{28.0}\\&= 71.43\ \mathrm{mol}\end{aligned}\]
\[\begin{aligned}n(H_2) &= \frac{1000}{2.00}\\&= 500\ \mathrm{mol}\end{aligned}\]
\[\begin{aligned}n(H_2)_{required} &= 3(71.43)\\&= 214.29\ \mathrm{mol}\end{aligned}\]
\[\begin{aligned}n(NH_3) &= 2(71.43)\\&= 142.86\ \mathrm{mol}\end{aligned}\]
\[\begin{aligned}m(NH_3) &= 142.86(17.0)\\&= 2.43\times10^3\ \mathrm g\end{aligned}\]
\[\begin{aligned}m(H_2)_{left} &= (500-214.29)(2.00)\\&= 571\ \mathrm g\end{aligned}\]
Final answers by part(a)\[2.43\times10^3\ \mathrm g\ NH_3
(b)\quad H_2\text{ is excess and }571\ \mathrm g\text{ remains}\]