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204 Chapter 4 Stoichiometry of Chemical Reactions The stoichiometric Si:N2 ratio is: 0.0712 mol Si = 1.33 mol Si 0.0535 mol N2 1 mol N2 3 mol Si = 1.5 mol Si 2 mol N2 1 mol N2 Comparing these ratios shows that Si is provided in a less-than-stoichiometric amount, and so is the limiting reactant. Alternatively, compute the amount of product expected for complete reaction of each of the provided reactants. The 0.0712 moles of silicon would yield mol Si3 N4 produced = 0.0712 mol Si × 1 mol Si3 N4 = 0.0237 mol Si3 N4 3 mol Si while the 0.0535 moles of nitrogen would produce mol Si3 N4 produced = 0.0535 mol N2 × 1 mol Si3 N4 = 0.0268 mol Si3 N4 2 mol N2 Since silicon yields the lesser amount of product, it is the limiting reactant. Check Your Learning Which is the limiting reactant when 5.00 g of H2 and 10.0 g of O2 react and form water? Answer: O2 Percent Yield The amount of product that may be produced by a reaction under specified conditions, as calculated per the stoichiometry of an appropriate balanced chemical equation, is called the theoretical yield of the reaction. In practice, the amount of product obtained is called the actual yield, and it is often less than the theoretical yield for a number of reasons. Some reactions are inherently inefficient, being accompanied by side reactions that generate other products. Others are, by nature, incomplete (consider the partial reactions of weak acids and bases discussed earlier in this chapter). Some products are difficult to collect without some loss, and so less than perfect recovery will reduce the actual yield. The extent to which a reaction’s theoretical yield is achieved is commonly expressed as its percent yield: percent yield = actual yield × 100% theoretical yield Actual and theoretical yields may be expressed as masses or molar amounts (or any other appropriate property; e.g., volume, if the product is a gas). As long as both yields are expressed using the same units, these units will cancel when percent yield is calculated. Example 4.13 Calculation of Percent Yield Upon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal was obtained according to the equation: What is the percent yield? Solution CuSO4(aq) + Zn(s) ⟶ Cu(s) + ZnSO4(aq) This content is available for free at http://cnx.org/content/col11760/1.9PDF Image | Stoichiometry of Chemical Reactions
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