Dinitrogen and dihydrogen react with each other to produce ammonia according to the following equation : N_2(g) + 3H_2(g) to 2NH_3(g) (i) Calculate the mass of ammonia produced if 2.00 xx 10^3 g dinitrogen react with 1.00 xx 10^(3)g of dihydrogen. (ii) Will any of the two reactants remain unreacted? (iii) If yes, which one and what would be its mass?

Dinitrogen and dihydrogen react with each other to produce ammonia acc

1.	Dinitrogen and dihydrogen react with each other to produce ammonia according to the following equation : N_2(g) + 3H_2(g) to 2NH_3(g) (i) Calculate the mass of ammonia produced if 2.00 xx 10^3 g dinitrogen react with 1.00 xx 10^(3)g of dihydrogen. (ii) Will any of the two reactants remain unreacted? (iii) If yes, which one and what would be its mass?
Answer» Solution :(i) `underset("1 MOLE 28 g")(N_(2)) + underset("3 MOLES of" 3 xx 2= 6g)(3H_(2)) to underset("2 moles" 2 xx 17 = 34 g)(2NH_(3))` `therefore` 28 g of `N_(2)` react with hydrogen = 6g `therefore 2.00 xx 10^(3) g` of `N_(2)` react with hydrogen `=6/28 xx 2.00 xx 10^(3) = 428.57 g` OBVIOUSLY, dihydrogen is in excess and dinitrogen is the LIMITING REAGENT, v 28 g of dinitrogen produce ammonia = 34 g `therefore 2.00 xx 10^3` g of dinitrogen will produce ammonia = `34/28 xx 2.00 xx 10^(3) = 2428.57 g` (ii) Dihydrogen will remain unreacted. (iii) Mass of `H_2` left unreacted = `1.00 xx 10^(3) - 428.57 = 571.43 g`