Find the limiting and excess reagent in the following cases :- A. N2 ( 100g ) + 3h2( 100g) ------> 2NH3 B. C2H2 ( 125g) + H2 ( 50g ) ------> C2H4 Share with your friends Share 0 Aditi Gupta answered this Solution: The reaction is as follows- A) N2 + 3H2 → 2NH3Moles of N2 = 10028 = 3.57 molMoles of H2 = 1002 = 50 molNow we have to divide moles with the reaction coefficient.For N2 reaction coefficient is 1.For H2 reaction coefficient is 3.So, 503 = 16.67Still moles of N2 is less, so nitrogen is the limiting reagent and hydrogen is the excess reagent.B) C2H2 + H2 → C2H4Moles of C2H2 = 12526 = 4.80Moles of H2 = 502 = 25Both have the reaction coefficient as 1, so here the limiting reagent is C2H2 because number of moles is less and the excess reagent is hydrogen. 0 View Full Answer
