# 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?

## 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]

```