1. The problem statement, all variables and given/known data Ammonia reacts with O2 to form either NO(g) or NO2(g) according to these unbalanced equations: NH3 + O2 ----> NO + H2O NH3 + O2 ----> NO2 + H2O In a certain experiment 2.00 moles of NH3 and 10.00 moles of O2 are contained in a closed flask. After the reaction is complete, 6.75 moles of O2 remains. Calculate the number of moles of NO in the product mixture: (Hint: You cannot do this problem by adding the balanced equations because you cannot assume that the two reactions will occur with equal probability.) 2. Relevant equations 3. The attempt at a solution After the balance of equations, x=mol of NH3 in the first equations(=mol of NO as well) 1st reaction: 2 mol NH3 gives 5/2 mole O2 x mol gives ... 5x/4 mole O2 2nd reaction: 2 mol NH3 gives 7/2 mole O2 (2-x) mol gives ... 7(2-x)/4 mole O2 5x/4 + 7(2-x)/4 + 6.75=10 I find: x=0.5 mole of NH3 as well as 0.5 mole of NO But solution manual has 0.48 as answer, is my calculation wrong in any way??