not), nA the number of moles (n) of substance A. In a fed-batch reactor some reactants/ingredients are added continuously or in pulses (compare making porridge by either first blending all ingredients and then letting it boil, which can be described as a batch reactor, or by first mixing only water and salt and making that boil before the other ingredients are added, which can be described as a fed-batch reactor). Mass balances for fed-batch reactors become a bit more complicated. Reactive example In the first example, we will show how to use a mass balance to derive a relationship between the percent excess air for the combustion of a hydrocarbon-base fuel oil and the percent oxygen in the combustion product gas. First, normal dry air contains of oxygen per mole of air, so there is one mole of in of dry air. For stoichiometric combustion, the relationships between the mass of air and the mass of each combustible element in a fuel oil are: Considering the accuracy of typical analytical procedures, an equation for the mass of air per mass of fuel at stoichiometric combustion is: where wC, wH, wS, and wO refer to the mass fraction of each element in the fuel oil, sulfur burning to SO2, and AFRmass refers to the air-fuel ratio in mass units. For of fuel oil containing 86.1% C, 13.6% H, 0.2% O, and 0.1% S the stoichiometric mass of air is , so AFR = 14.56. The combustion product mass is then . At exact stoichiometry, should be absent. At 15 percent excess air, the AFR = 16.75, and the mass of the combustion product gas is , which contains of excess oxygen. The combustion gas thus contains 2.84 percent by mass. The relationships between percent excess air and % in the combustion gas are accurately expressed by quadratic equations, valid over the range 0–30 percent excess air: In the second example, we will use the law of mass action to derive the expression for a chemical equilibrium constant. Assume we have a closed reactor in which the following liquid phase reversible reaction occurs: The mass balance for substance A becomes As we have a liquid phase reaction we can (usually) assume a constant volume and since we get or In many textbooks this is given as the definition of reaction rate without specifying the implicit assumption that we are talking about reaction