Question

Bonus Question (20 points). Nitrogen dioxide gas reacts with water vapor to produce oxygen and ammonia gases. Suppose that 12.8 g of nitrogen dioxide gas reacts with 5.00 L of water vapor 375 °C and 725 tor? how many moles of nitrogen dioxide gas are there initially? b. how many moles of water vapor are there initially? e Balance the chemical reaction. Complete the following table after solving and NO + H2O - O d. What is the limiting reagent? Determine the liming reagent by comparing the theoretical mole ratio to the actual me Theoretical mole ratio Actual mole ratio Limiting reagent e. what does x equal? f. how many grams of ammonia can be produced? Page 7 of 11

Answer 1

The balanced equation is 4 NO2 + 6 H20 ---------> 4 NH3 + 7 02 (a) Number of moles of NO2 gas = mass/mol.wt. = 12.8 g/(46.01 g/mol) = 0.2782 mol (b) Number of moles of H20 = n = PV/RT = (725/760) atm * 5.00 L/(0.0821 L.atm.mol-1.6-1. * 648.15 K = 0.089635 mol mo - 4 NO2 + 0.2782 - 4 x 0.2782 - 4x 6 H2O ---------> 0.089635 - 6x 0.089635 - 6x 4 NH3 + 702 0 0 + 4x + 7x 4x 7x (d) Theoretical mole ratio: NO2: H20 = 4:6 = 2:3 Actual Mole ratio: 0.2782 : 0.089635 = 3.10 : 1 The number of moles of water should be more than that of NO2 But the number of moles of H20 is less than that of NO2. So, H20 is the Limiting reagent. (e) 6x = 0.089635 X = 0.014939 Now, the ICE Table is (c) 4 NH3 + 7 02 4NO2 + 0.2782 -0.059756 0.2184 6 H20 ---------> 0.089635 -0.089635 +0.059756 0.059756 +0.104574 0.104574 (f) Number of moles of NH3 formed = 0.059756 mol Mass of NH3 = moles * mol.wt. = 0.059756 mol * 17.03 g/mol = 1.02 g