Mole ratio with limiting reactant

Complex stoichiometry problems should be worked slowly and carefully, one step at a time. When solving a problem that deals with limiting reactants, the idea is to find how many moles of all reactants are actually present and then compare the mole ratios of those actual amounts to the mole ratios required by the balanced equation. That comparison will identify the reactant there is too much of (the excess reactant) and the reactant there is too little of (the limiting reactant). 

Now we must determine which of the three reactants is limiting. To do so, we will compare the mole ratios of reactants required by the balanced equation with the actual mole ratios. For the reactants K2C03 and FeCr204 the required mole ratio is... 

Since we cannot count individual particles, we again must turn to our macroscopic-particulate link, the mole. The amount of product is limited by the number of moles of the limiting reactant, and it must be calculated from that number of moles. If we start with three moles of carbon and two moles of oxygen, we have the same ratio of particles as when we start with three carbon atoms and two oxygen molecules. If two moles of the limiting reactant—oxygen—react, two moles of carbon dioxide are produced. According to the equation, each mole of carbon dioxide produced requires one mole of carbon to react. 

Make a table like the one above and determine the number of moles of acetic acid (HAc) and acetate ion (Ac-) after the reaction is complete. Since the stoichiometric ratios are 1 1, the limiting reactant is the one with the smaller number of moles. 

When two substances react, they react in exact amounts. You can determine what amounts of the two reactants are needed to react completely with each other by means of mole ratios based on the balanced chemical equation for the reaction. In the laboratory, precise amounts of the reactants are rarely used in a reaction. Usually, there is an excess of one of the reactants. As soon as the other reactant is used up, the reaction stops. The reactant that is used up is called the limiting reactant. Based on the quantities of each reactant and the balanced chemical equation, you can predict which substance in a reaction is the limiting reactant. 

Note especially that the number of moles of Nal exceeds the number of moles of Pb(N03)2 present, but that these numbers are not what must be compared. Compare the number of moles of one reactant present with the number of moles of that same reactant required The ratio of moles of Nal to Pb(N03)2 in the equation is 2 1, but in the sample that ratio is less than 2 1 therefore, the Nal is in limiting quantity. 

Positive Slope The mL and mole ratio for equivalency is 4 NaClO to 1 Na2S203 (see 1 above). In examining the data table, it can be seen that the Na2S203 is in excess, with the NaClO being the limiting reactant for example, in Trial 7, for 30 mL of 0.500 M NaClO,... 

Which reactant is the limiting reactant How does the experimental mole ratio of Fe to Cu compare with the mole ratio in the balanced chemical equation What is the percent yield ... 

For SK-500 the rate at 573°K and 400 sec after the initiation of reactant flow is independent of reactant mole ratio for Ce C2 = 0.7 to 10. Under these conditions the 400-sec point is just beyond the maximum in the rate curve. Similar behavior was observed at one other condition. Initial rate of reaction estimated by extrapolating the decay portion of the rate curves for this data to zero time (see below) indicates a maximum in the rate at C6 C2 == 3.5 (Figure 2). Error bars represent estimated 95% confidence limits. The observed activity for HY is about twice that of SK-500, that for LaY is about two-thirds that of SK-500 (Figure 2). This is consistent with the trend expected (7) since all catalysts were activated to the same temperature. The temperature dependence of the observed rate is large for all systems studied indicating the absence of external mass transfer limitations. 

From the balanced equation, you can see that it is necessary to have twice as many moles of potassium iodide as lead nitrate for the reaction to complete. What if you only had 1.5 moles of KI, and 1 mole of Pb(N03)2 Since the ratio of KI to Pb(N03)2 is 2-to-l, if you only have 1.5 moles of KI, you could only combine with 0.75 moles of Pb(N03)2. Therefore, you would have 0.25 moles of unreacted Pb(N03)2 left in the reaction vessel after all KI had reacted. In this example, KI was the limiting reactant. 

A related, but simpler, way to determine which reactant is limiting is to compare the mole ratio of the substances required by the balanced equation with the mole ratio of reactants actually present. For example, in this case the mole ratio of H2 to N2 required by the balanced equation is... 

Suppose (nA)feed is the number of moles of an excess reactant. A, present in the feed to a reactor and that (nA)stoich lit stoichiometric requirement of A, or the amount needed to react completely with the limiting reactant. Then (nA)feed ("A)stoich s the amount by which the A in the feed exceeds the amount needed to react completely if the reaction goes to completion. The fractional excess of the reactant is the ratio of the excess to the stoichiometric requirement ... 

As a shortcut to determining the limiting reactant, all you have to do is to calculate the mole ratio(s) of the reactants and compare each ratio with the corresponding ratio of the coefficients of the reactants in the chemical equation thus ... 

We see that each mole of O2 requires exactly 0.500 mol of CH4 to be completely used up. We have 0.667 mol of CH4 for each mole of O2, so there is more than enough CH4 to react with the O2 present. That means that there is insufficient O2 to react with all of the available CH4. The reaction must stop when the Oj is gone Oj is the limiting reactant, and we must base the calculation on it. (If the available ratio of CH4 to O2 had been smaller than the required ratio, we would have concluded that there is not enough CH4 to react with all of the O2, and CH4 would have been the limiting reactant.)... 

The molar ratio of the reactants is 1 Hg(N03)2/l Na2S. Therefore, Hg(N03)2 is limiting because there are fewer moles of it than are needed to react with the moles of NajS. Finding grams of product from moles of limiting reactant and the molar ratio ... 

Because 7.72 mol of O2 is available but only 5 mol is needed to react with 1 mol of Pa, O2 is in excess and P4 is the limiting reactant. Use the moles of Pa to determine the moles of PaOio that will be produced. Multiply the number of moles of P4 by the mole ratio of PaOio [the unknown) to P4 [the known). 

To determine the limiting reactant, the actual mole ratio of the available reactants must be compared with the ratio of the reactants obtained from the coefficients in the balanced chemical equation. 

Plan To identify the hmiting reactant, we can calculate the number of moles of each reactant and compare their ratio with the ratio of coefficients in the balanced equation. We then use the quantity of the limiting reactant to calculate the mass of water that forms. 

All stoichiometry problems can be approached with this general pattern. In some cases, however, additional calculations may be needed. For example, if we are given (or able to measure) known amounts of two or more reactants, we must determine which of them will be completely consumed (the limiting reactant.) Once again, the mole ratios in the balanced equation hold the key. Another type of calculation that can be considered in a stoichiometry problem is determining the percentage yield of a reaction. In this case, the amount of product determined in the problem... 

Calculate the mole ratio of Sn to I2 used in the procedure. Which is the limiting reagent Why is the reaction carried out with this ratio of reactants and under a nitrogen atmosphere ... 

If we use a pH jiunp, as an example for the consequence of a temperature perturbation of an equilibrium, one can easily derive from the above that, in a solution in Tris buffer (pAT= 8 and AH=45.6 kJ mole ), an increase of 5 degrees will reduce the pATby 0.12. The corresponding shift of a solution at pH 8 would be to pH 7.88. The effect of a temperature jump, or of any other perturbation of the equilibrium constant, is maximal when the ratio of the reactants is unity. In a simple first order isomerization one has little control over this, but most of the reactions under discussion are, at least distantly, linked to a pseudo first order process with equilibrium positions dependent on reactant concentrations. Unfortunately the choice of experimental conditions (concentrations) is often limited by solubilities and the optical properties used for monitoring the reactants, a problem also encountered in... 

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