Ayield of nh3 of approximately 98% can be obtained at 200°c and 1,000 atmospheres of pressure. how many grams of n2 must react to form 1.7 grams of ammonia, nh3? 0.0058 g .052 g 1.4 g 2.8 g

Since the percent yield of the reaction is 98%, that means that 1.7g of ammonia is 98% of the theoretical yield so you need to find what the theoretical yield is by dividing 1.7 by .98 to get 1.735g.  Since you know that the theoretical yield is 1.735g you can use stoichiometry to figure out how much N₂ is required in order to produce 1.735g of ammonia.  

To do that you first need to find the balanced chemical equation for the reaction which is N₂+3H₂⇒2NH₃.  Then you need to find how many moles of ammonia are in 1.735g by dividing 1.735 by the molar mass of ammonia (17g/mol) to get 0.1021mol of ammonia.  After that you can find the number of moles of N₂ required to make 0.1021mol of ammonia by using the fact that 1mol N₂ is used to make 2mol NH₃ so you divide 0.1021 by 2 to get 0.05103mol N₂.  lastly you can find the number of grams of N₂ by multiplying 0.05103 by the molar mass of N₂ (28g/mol) to get 1.42g of N₂.  Therefore 1.4g of N₂ is required to make 1.7g of ammonia.

I hope this helps. Let me know in the comments if anything is unclear.