J.R. S. answered • 02/22/20
Ph.D. University Professor with 10+ years Tutoring Experience
4NH3(g) + 5O2(g) ==> 4NO(g) + 6H2O(g) ... balanced equation
This is a limiting reactant type of question, so we must first determine which reactant (NH3 or O2) is in limiting supply. There are essentially 2 ways to go about this. I'll present both so you can choose which you prefer.
Method 1: Determine moles of each reactant and then divide that number by the coefficient in the balanced equation...
*** For NH3 we have 61.7 g NH3 x 1 mol NH3/17.03 g = 3.62 mol (divided by 4 = 0.906 mol)
*** For O2 we have 61.7 g O2 x 1 mol/32 g = 1.93 mol (divided by 5 = 0.385 mol)
This shows that O2 is limiting since 0.385 is less than 0.906.
Now, to find the maximum mass H2O formed, we use the ORIGINAL moles of O2 that we determined, i.e. we us the 1.93 moles (do not use the 0.385 moles).
*** 1.93 moles O2 x 6 mol H2O/5 mol O2 x 18 g H2O/mol H2O = 41.7 g H2O
Method 2: Determine mass of water produced from each reactant and which ever is less, that is the limiting reactant and the answer will be the maximum mass that can be produced...
*** For NH3 we have 61.7 g NH3 x 1 mol/17.0 g x 6 mol H2O/4 mol NH3 x 18 g H2O/mol = 98.0 g H2O
*** For O2 we have 61.7 g O2 x 1 mol/32 g x 6 mol H2O/5 mol O2 x 18 g H2O/mol = 41.7 g H2O
