Question

5. Ammonia reacts with O, to form either NO(g) or NO, (g) according to these unbalanced equations: 4 NH, (g) + 50, (g) → NO(g) +_U HO(g) Y NH, (g) + _302(g) → _4 NO, (g) + 6H2O(g) a. In a certain experiment 2.00 moles of NH, (g) and 10.00 moles of Oz (g) are contained in a closed flask. After the reaction is complete, 6.75 moles of O, (g) remains. Calculate the number of moles of NO (g) in the product mixture.

Answer 1

on to produce Ammonia No rearts with and H2O Step 1: NH3(g) + O2(g) → NO(g) + H2O(g) [ NOT Balanced] Step: Balonced - 4 NH3(g) +502 (g) 4 NOCg) + 6H₂O(g) [ Balanced equation] 6. San given 2 mole Wholg) and to moles o2g remaing O2 in 6.75 moles. we have to find moles of No ? 4 NH₃(g) +502 (g) - 4 NO + 6H2016 (92 intially 2 moles To moles. 2- 4x to.se 4 x 6x at equitim VAR

at equilistium 6.75 mole moles of on left is to-5x = 6.75 S x = 3.25 x=0.65moly .: In the above equation at equilitum moles of no ag) = 4x - 4x0.65 moles of Norg) * 2.6 moles