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In the examples above, one reactant was present in excess One reactant was completely consumed, and some of the other reactant would be left over The reactant that is used up first is called the limiting reactant (LR) This reactant really determines the amount of product being formed How is the limiting reactant determined You can t assume it is the reactant in the smallest amount, since the reaction stoichiometry must be considered There are generally two ways to determine which reactant is the limiting reactant: 1 Each reactant, in turn, is assumed to be the limiting reactant, and the amount of product that would be formed is calculated The reactant that yields the smallest amount of product is the limiting reactant The advantage of this method is that you get to practice your calculation skills; the disadvantage is that you have to do more calculations 2 The moles of reactant per coefficient of that reactant in the balanced chemical equation is calculated The reactant that has the smallest mole-to-coefficient ratio is the limiting reactant This is the method that many use Let us consider the Haber reaction once more Suppose that 500 g of nitrogen and 400 g of hydrogen were allowed to react Calculate the number of grams of ammonia that could be formed First, write the balanced chemical equation: N2(g) + 3 H2(g) 2 NH3(g) Next, convert the grams of each reactant to moles: 500 g N 2 1 mol N 2 = 17848 mol N 2 1 28014 g N 2 400 g H 2 1 mol H 2 = 198432 mol H 2 1 20158 g H 2 Divide each by the coefficient in the balanced chemical equation The smaller is the limiting reactant: For N2: 17848 mol N2/1 = 17848 mol/coefficient limiting reactant For H2: 198432 mol H2/3 = 66144 mol/coefficient Finally, base the stoichiometry of the reaction on the limiting reactant: 2 mol NH3 170307 g 500 g N 2 1 mol N 2 = 608 g NH3 1 28014 g N 2 1 mol N 2 1 mol NH3.

refers to the vapor Superscripts v and 1 are not used here because of a presumption that phase and that y; and f , are liquid-phase properties Substituting for f , by Eq (1141) gives:

There are some people I knew I couldn t push Some people I had my doubts about, and I pushed them and berated them to find out what I could about their character, their limits Those are the things that are important to me, because this is what the game is all about

Since the Poynting factor (represented by the exponential) at low to moderate pressures differs from unity by only a few parts per thousand, its omission introduces negligible error, and this eqJation is often simplified:

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Equation (141), called the gammalphi formulation of VLE, reduces to Raoult's law when Qi = yi = 1, and to modified Raoult's law when Qi = 1 Systematic application of Eqs (141) and (142) depends on the availability of correlations of data from which values may be obtained for P isat, Qi, and yi The vapor pressure of a pure species is most commonly given by the Antoine equation, Eq (671):

Restriction to moderate pressures allows calculation of the fugacity coefficients in Eq (142) to be based on Eq (337), the two-term virial expansion in P They are then given by Eq (1 161), here written:

Anytime the quantities of more than one reactant are given it is probably a LR problem Let s consider another case To carry out the following reaction: P2O5(s)+3H2O(l) 2H3PO4 (aq) 125 g of P2O5 and 500 g of H2O were supplied How many grams of H3PO4 may be produced

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