The equation describing the reaction between ammonia and oxygen is described as follows:
4 NH3 + 5 O2 ...........> 4 NO + 6 H2O
From the periodic table:
mass of oxygen = 16 grams
mass of hydrogen = 1 gram
mass of nitrogen = 14 grams
Therefore,
mass of NH3 = 14 * 3(1) = 17 grams
mass of O2 = 2*16 = 32 grams
mass of H2O = 2(1) + 16 = 18 grams
From the balanced equation above:
4(17) = 68 grams of NH3 are required to react with 160 grams of oxygen to produce 108 grams of water
In this reaction, NH3 is the limiting reagent. This means that the reaction would stop once the amount of NH3 is consumed.
Therefore, we will base the calculations on the amount of NH3.
Now, we know that 68 grams of ammonia produce 108 grams of water. To know the amount of water produced from 54.5 grams of NH3, we will simply do a cross multiplication as follows:
amount of water = (54.5*108) / 68 = 86.5588 grams
