zHW: Combusion of Ammonia in Air
Ammonia, NH3, is burned in air, produces nitrogen and water by the unbalanced reaction
Given that you started with 51.0 g of NH3, how many g of water will be produced?
Contents
Prepare list of Species
As usual for these problems, begin by specifying the list of species involved in the reaction. We record the position of ammonia and water in this list using their chemical formulas as Matlab variables.
species = {'NH3','O2','N2','H2O'}; % Record for later use the position of NH3 and H2O in this list NH3 = 1; H2O = 4; molweight(species);
Species Mol. Wt. ------- -------- NH3 17.03 O2 32.00 N2 28.01 H2O 18.02
Balance Reaction
The balanced reaction is computed using stoich and displayed using disp_reaction. disp_reaction returns integer stoichiometric coefficients, which we use for subsequent calculations.
V = stoich(species); V = disp_reaction(V,species);
Convert Starting Mass of Ammonia to Moles
The the mass of ammonia to moles.
nNH3 = 51.0/molweight('NH3'); fprintf('\nStarting amount of Ammonia = %g [moles]\n',nNH3);
Starting amount of Ammonia = 2.99462 [moles]
Calculate Moles of Water Produced
The key concept behind this problem is that consuming V(H20) moles of water are produced for every -V(NH3) moles of ammonia reacted.
nH2O = - (V(H2O)/V(NH3))*nNH3;
fprintf('\nWater Produced = %g [moles]\n',nH2O);
Water Produced = 4.49193 [moles]
Convert Moles of Water to Mass
fprintf('\nMass of Water Produced = %g [g]\n\n', nH2O*molweight('H2O'));
Mass of Water Produced = 80.9233 [g]