My resources follow the New AP Chemistry Course Framework.This worksheet has 45 multiple choice questions on the following topics of Unit 4: Chemical ReactionsUnit 4.5: Reaction StoichiometryReading &, This Printable AP Chemistry Worksheet contains sets of carefully selected high-quality multiple choice questions on Reaction Stoichiometry. The second equation also has a gram-mole limiting reagent question. endobj
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138 C 7 H 6 O 3 1 mol C 7 H 6 O 3 1 mol C 9 H 8 O 4, 4 g C 4 H 6 O 3 x 1mol C 4 H 6 O 3 x 1 mol C 9 H 8 O 4 x 180 g C 9 H 8 O 4 = 7 g C 9 H 8 O 4 13 0 obj
Limiting Reactant Worksheet Answers limiting theoretical and percentage yields key ko2 h2o koh (aq) o2 if reaction vessel contains 0.15 mol ko2 and 0.10 mol h2o Skip to document Ask an Expert Sign inRegister Sign inRegister Home Ask an ExpertNew My Library Discovery Institutions Silver Creek High School (Colorado) University of Georgia Because the reactants both have coefficients of 1 in the balanced chemical equation, the mole ratio is 1:1. 7.2 Limiting Reagent and Reaction Yields Learning Objectives By the end of this section, you will be able to: Explain the concepts of theoretical yield and limiting reactants/reagents. We can replace mass by the product of the density and the volume to calculate the number of moles of each substance in 10.0 mL (remember, 1 mL = 1 cm3): \[ \begin{align*} \text{moles} \; \ce{C2H5OH} & = { \text{mass} \; \ce{C2H5OH} \over \text{molar mass } \; \ce{C2H5OH} }\nonumber \\[6pt] & = {( \text{volume} \; \ce{C2H5OH} ) \times (\text{density} \, \ce{C2H5OH}) \over \text{molar mass } \; \ce{C2H5OH}}\nonumber \\[6pt] &= 10.0 \, \cancel{ml} \; \ce{C2H5OH} \times {0.7893 \, \cancel{g} \; \ce{C2H5OH} \over 1 \, \cancel{ml} \, \ce{C2H5OH} } \times {1 \, mol \; \ce{C2H5OH} \over 46.07 \, \cancel{g}\; \ce{C2H5OH}}\nonumber \\[6pt] &= 0.171 \, mol \; \ce{C2H5OH} \\[6pt] \text{moles} \; \ce{CH3CO2H} &= {\text{mass} \; \ce{CH3CO2H} \over \text{molar mass} \, \ce{CH3CO2H}}\nonumber \\[6pt] &= { (\text{volume} \; \ce{CH3CO2H} )\times (\text{density} \; \ce{CH3CO2H}) \over \text{molar mass} \, \ce{CH3CO2H}}\nonumber \\[6pt] &= 10.0 \, \cancel{ml} \; \ce{CH3CO2H} \times {1.0492 \, \cancel{g} \; \ce{CH3CO2H} \over 1 \, \cancel{ml} \; \ce{CH3CO2H}} \times {1 \, mol \; \ce{CH3CO2H} \over 60.05 \, \cancel{g} \; \ce{CH3CO2H} } \\[6pt] &= 0.175 \, mol \; \ce{CH3CO2H}\nonumber \end{align*} \nonumber \]. !+PN0gS2f9xkwTKEIN%MJtX@P 102 g C 4 H 6 O 3 1 mol C 4 H 6 O 3 1 mol C 9 H 8 O 4, Percent yield aspirin = 2 g C 9 H 8 O 4 x 100 = 84 % yield aspirin Although titanium is the ninth most common element in Earths crust, it is relatively difficult to extract from its ores. PDF. 20 0 obj
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d. Each chemical equation comes with 2 limiting reagent calculations and one percent yield question. MV#O]G` 8Y
as isolated $rom the ,rodcts' hat as the ,recentage /ield o$, enene ith chlorine and to e&,ect a /ield no higher that 05@. More often, however, reactants are present in mole ratios that are not the same as the ratio of the coefficients in the balanced chemical equation. Legal. Limiting Reagents and Percentage Yield Worksheet 1. Consequently, none of the reactants was left over at the end of the reaction. %
The total number of moles of Cr2O72 in a 3.0 mL Breathalyzer ampul is thus, \[ moles\: Cr_2 O_7^{2-} = \left( \dfrac{8 .5 \times 10^{-7}\: mol} {1\: \cancel{mL}} \right) ( 3 .0\: \cancel{mL} ) = 2 .6 \times 10^{-6}\: mol\: Cr_2 O_7^{2}\nonumber \], C The balanced chemical equation tells us that 3 mol of C2H5OH is needed to consume 2 mol of \(\ce{Cr2O7^{2}}\) ion, so the total number of moles of C2H5OH required for complete reaction is, \[ moles\: of\: \ce{C2H5OH} = ( 2.6 \times 10 ^{-6}\: \cancel{mol\: \ce{Cr2O7^{2-}}} ) \left( \dfrac{3\: mol\: \ce{C2H5OH}} {2\: \cancel{mol\: \ce{Cr2O7^{2 -}}}} \right) = 3 .9 \times 10 ^{-6}\: mol\: \ce{C2H5OH}\nonumber \]. Are the limiting reagents always completely consumed? The balanced equation for the reaction of iron (iii) phosphate . Here is some common terminology used to describe reactions based on the concentrations of reactions. 2. Review of balancing equations Web what is my percent yield? In this bundle I have:#1 Stoichiometry Test Review that contains mole to mole ratios, mole to mole conversions, molar mass calculations, mole to mass, mass to mole, mass to mass, limiting reagent and percent yield.#2 Stoichiometry Quiz. endobj
yield of AlCl3 of just 135.5 grams, the percent yield would be 72.04%. i) what mass of iodine was produced? Because it is also highly resistant to corrosion and can withstand extreme temperatures, titanium has many applications in the aerospace industry. Limiting reactant. The first problem is a real life situation about baking cookies then it moves into two simple problems to practice with the equation. 4di[h`NAZ?e0Is=ir'QSGzFAiMsj5 reactant? Another Limiting Reagent Worksheet: Part two of the limiting reagent saga. Percent Yield Calculations: Using theoretical and actual yields to determine whether the reaction was a success. bed k$pF`S.fw endstream
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Limiting Reagent Worksheets 1. Consider the reaction : I 2 O 5 (g) + CO (g) CO 2 (g) + I 2 (g) [A] 80.0 grams of iodine (V) oxide, I 2 O 5 , reacts with 28.0 grams of CO. endobj
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2015 AP Chemistry free response 2a (part 1 of 2) %PDF-1.5
Add highlights, virtual manipulatives, and more. This is a review worksheet for students to practice and assess their knowledge of stoichiometry (mass-mass, volume-volume, limiting reagent, and percent yield). unconsumed? A 100% yield means that everything worked perfectly, and the chemist obtained all the product that could have been produced. You should contact him if you have any concerns. { "7.01:_Stoichiometric_Reactions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "7.02:_Theoretical_Yield_Limiting_and_Excess_Reagents" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "7.03:_Percent_Yield" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "00:_Front_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "00:_General_Information" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "01:_Introduction" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "02:_Mathematical_Fundamentals" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "03:_Atoms_and_Elements" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "04:_Compounds_and_Molecules" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "05:_Chemical_Reactions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "06:_Counting_Molecules_through_Measurements" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "07:_Stoichiometry" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "08:_Solution_Chemistry" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "09:_Gases" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "10:_Thermodynamics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "zz:_Back_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, 7.2: Theoretical Yield, Limiting and Excess Reagents, https://chem.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fchem.libretexts.org%2FCourses%2FUniversity_of_Arkansas_Little_Rock%2FChem_1300%253A_Preparatory_Chemistry%2FLearning_Modules%2F07%253A_Stoichiometry%2F7.02%253A_Theoretical_Yield_Limiting_and_Excess_Reagents, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), Stoichiometric Proportions and Theoretical Yield, status page at https://status.libretexts.org, Understanding Limiting and Excess Reagents, Predict quantities of products produced or reactants consumed based on complete consumption of limiting reagent (on both mole and mass basis). Convert the number of moles of product to mass of product. Note in the video how we first wrote the balanced equation, and then under each species wrote down what we were given. xYmkGn7w%NPRCI?%t^H;H{*ig^7o^z{oHWrO5UOS This metal is fairly light (45% lighter than steel and only 60% heavier than aluminum) and has great mechanical strength (as strong as steel and twice as strong as aluminum). Consider the reaction I2O5(g) + 5 CO(g) -------> 5 CO2(g) + I2(g) a) 80.0 grams of iodine(V) oxide, I2O5, reacts with 28.0 grams of carbon monoxide, CO. Calculate the percent yield for a reaction. %%EOF
Quizzes with auto-grading, and real-time student data. For example, there are 8.23 mol of \(\ce{Mg}\), so (8.23 2) = 4.12 mol of \(\ce{TiCl4}\) are required for complete reaction. ? 9 0 obj
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percent yield of this reaction? The number of moles of each is calculated as follows: \[ \begin{align} \text{moles} \; \ce{TiCl4} &= \dfrac{\text{mass} \, \ce{TiCl4}}{\text{molar mass} \, \ce{TiCl4}}\nonumber \\[4pt] &= 1000 \, \cancel{g} \; \ce{TiCl4} \times {1 \, mol \; TiCl_4 \over 189.679 \, \cancel{g} \; \ce{TiCl4}}\nonumber \\[4pt] &= 5.272 \, mol \; \ce{TiCl4} \\[4pt] \text{moles }\, \ce{Mg} &= {\text{mass }\, \ce{Mg} \over \text{molar mass }\, \ce{Mg}}\nonumber \\[4pt] &= 200 \, \cancel{g} \; \ce{Mg} \times {1 \; mol \, \ce{Mg} \over 24.305 \, \cancel{g} \; \ce{Mg} }\nonumber \\[4pt] &= 8.23 \, \text{mol} \; \ce{Mg} \end{align}\nonumber \]. Given the following reaction: (Balance the equation first!) Web web limiting reagents and percentage yield worksheet 1 consider the reaction i2o5 g 5 co g 5 co2 g i2 g a. MsRazz ChemClass. The first step is to calculate the number of moles of each reactant in the specified volumes: \[ moles\: K_2 Cr_2 O_7 = 500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .17\: mol\: K_2 Cr_2 O_7} {1\: \cancel{L}} \right) = 0 .085\: mol\: K_2 Cr_2 O_7\nonumber \], \[ moles\: AgNO_3 = 250\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .57\: mol\: AgNO_3} {1\: \cancel{L}} \right) = 0 .14\: mol\: AgNO_3\nonumber \]. HMk1aasxp=V Derive the theoretical yield for a reaction under specified conditions. What is the Limiting Reagent and Theoretical Yield of Ag2S if 2.4 g Ag, 0.48 g H2S and 0.16g O2 react? Limiting reagent stoichiometry. This material has bothoriginal contributions, and contentbuilt upon prior contributions of the LibreTexts Community and other resources,including but not limited to: 7.2: Theoretical Yield, Limiting and Excess Reagents is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts. Therefore, the actual yield, the measured mass of products obtained from a reaction, is almost always less than the theoretical yield (often much less). Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. After identifying the limiting reactant, use mole ratios based on the number of moles of limiting reactant to determine the number of moles of product. Fe 2 O 3 (S) + 3 CO (g)! Are you getting the free resources, updates, and special offers we send out every week in our teacher newsletter? c. Determine the mass in grams of the product formed. Stoichiometric Proportions and Theoretical Yield The second page is a page to do with the students and the third page is a practice page students can do in class or for homework, that's up to you. Gravimetric analysis and precipitation gravimetry. Determine the mass of iodine I2, which could be produced? 40% 40% found this document not. easy limiting reagent worksheet all of the questions on this worksheet involve the following reaction: when copper (ii) chloride reacts with sodium nitrate, Skip to document Ask an Expert Sign inRegister Sign inRegister Home Ask an ExpertNew My Library Discovery Institutions Western Governors University Grand Canyon University University of Georgia Mole ratio If this is not the case, then the student must have made an error in weighing either the reactants or the products. <>
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Using mole ratios, determine which substance is the limiting reactant. Web any yield over 100% is a violation of the law of conservation of mass. 1. In this problem there are 3 reagents, and this technique allows us to quickly identify the, To calculate the excess reagent you determine how much is left over after the complete consumption of the limiting reagent, Massexcess reagent= Massinitial- Massconsumed by complete consumption of limiting reagent. B We need to calculate the number of moles of ethanol and acetic acid that are present in 10.0 mL of each. A Always begin by writing the balanced chemical equation for the reaction: \[ \ce{ C2H5OH (l) + CH3CO2H (aq) \rightarrow CH3CO2C2H5 (aq) + H2O (l)}\nonumber \]. \[ \text{theoretical yield of procaine} = 0.0729 \, mol \times {236.31 \, g \over 1 \, mol } = 17.2 \, g\nonumber \], C The actual yield was only 15.7 g of procaine, so the percent yield (via Equation \ref{3.7.3}) is, \[ \text{percent yield} = {15.7 \, g \over 17.2 \, g } \times 100 = 91.3 \%\nonumber \], (If the product were pure and dry, this yield would indicate very good lab technique!). 4.3: Limiting Reactant, Theoretical Yield, and Percent Yield is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts. Limiting Reactant Problems Using Molarities: Limiting Reactant Problems Using Molarities, YouTube(opens in new window) [youtu.be]eOXTliL-gNw (opens in new window). Modified from Limiting Reactant and Percent Yield Wkst.pdf Blake - 3/2015 STO.4 Solve stoichiometric problems from a balanced chemical equation. 345 0 obj
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Titanium is also used in medical implants and portable computer housings because it is light and resistant to corrosion. Introduction to Limiting Reactant Problems: Introduction to Limiting Reactant Problems, YouTube(opens in new window) [youtu.be]. In this worksheet, we will practice identifying the limiting reagent and calculating the percentage yield of desired products based on the actual and theoretical yield. Conversely, 5.272 mol of \(\ce{TiCl4}\) requires 2 5.272 = 10.54 mol of Mg, but there are only 8.23 mol. Determine the mass of I 2 , which could be produced? The second equation also has a gram-mole limiting reagent question. It occurs as concentrated deposits of a distinctive ore called galena (\(\ce{PbS}\)), which is easily converted to lead oxide (\(\ce{PbO}\)) in 100% yield by roasting in air via the following reaction: \[\ce{ 2PbS (s) + 3O2 \rightarrow 2PbO (s) + 2SO2 (g)}\nonumber \]. A stoichiometric quantity of a reactant is the amount necessary to react completely with the other reactant(s). Disclaimer: Some answers are in scientific notation or might not be included because I changed some of the questions from year to year. The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, rebelford@ualr.edu. In the first step of the extraction process, titanium-containing oxide minerals react with solid carbon and chlorine gas to form titanium tetrachloride (\(\ce{TiCl4}\)) and carbon dioxide. Calculate the number of moles of \(\ce{Cr2O7^{2}}\) ion in 1 mL of the Breathalyzer solution by dividing the mass of K. Find the total number of moles of \(\ce{Cr2O7^{2}}\) ion in the Breathalyzer ampul by multiplying the number of moles contained in 1 mL by the total volume of the Breathalyzer solution (3.0 mL). If 15 grams of copper (II) chloride react with 20 grams of sodium nitrate, how much sodium chloride Any reagents remaining after the complete consumption of the limiting reagent are know as excess reagents. Limiting Reagent and Percent Yield (mol-mol) Created by Robert Klaasen Two worksheets are included. . endobj
Legal. Displaying top 8 worksheets found for - Limiting And Excess Reactant. Check him out on YouTube first!This video guide pack comes with links to YouTube videos for each of the topics covered below:Stoichiometry Made Easy: Stoichiometry Tutorial 1Stoichiometry Made Easy: Stoichiometry Tutorial 2Stoichiometry Grams to Grams Tri, This lot includes all six of my dimensional analysis worksheets for middle school physical science or high school chemistry. When reactants are not present in stoichiometric quantities, the limiting reactant determines the maximum amount of product that can be formed from the reactants. If you have a dozen eggs, which ingredient will determine the number of batches of brownies that you can prepare? TPT empowers educators to teach at their best. Title: Limiting Reagent Worksheet Author: Moira O'Toole The reaction requires a 1:1 mole ratio of the two reactants, so p-aminobenzoic acid is the limiting reactant. Step 3: Because magnesium is the limiting reactant, the number of moles of magnesium determines the number of moles of titanium that can be formed: \[ mol \; \ce{Ti} = 8.23 \, mol \; \ce{Mg} = {1 \, mol \; \ce{Ti} \over 2 \, mol \; \ce{Mg}} = 4.12 \, mol \; \ce{Ti} \nonumber \] Thus only 4.12 mol of Ti can be formed. <>
As we saw in Example 1, there are many different ways to determine the limiting reactant, but they all involve using mole ratios from the balanced chemical equation. How to do a stoichiometric calculation 58 g NaCl 2 mol NaCl 1 mol HCl, 12 g H 2 SO 4 x 1 mol H 2 SO 4 x 2 mol HCl x 36 g HCl = 8 g HCl Consider the reaction I2O5 (g) + 5 CO (g) -------> 5 CO2 (g) + I2 (g) a) 80.0 grams of iodine (V) oxide, I2O5, reacts with 28.0 grams of carbon monoxide, CO. <>
The Breathalyzer is a portable device that measures the ethanol concentration in a persons breath, which is directly proportional to the blood alcohol level. Limiting Reagent Worksheet : There's no end to what you can achieve unless there's a limiting reagent involved. Thus 1.8 104 g or 0.18 mg of C2H5OH must be present. i. Step 3: calculate the mass carbon dioxide based on the complete consumption of the limiting reagent. It is prepared by reacting ethanol (\(\ce{C2H5OH}\)) with acetic acid (\(\ce{CH3CO2H}\)); the other product is water. If we are given the density of a substance, we can use it in stoichiometric calculations involving liquid reactants and/or products, as Example \(\PageIndex{1}\) demonstrates. Calculate the number of moles of each reactant present: 5.272 mol of \(\ce{TiCl4}\) and 8.23 mol of Mg. Divide the actual number of moles of each reactant by its stoichiometric coefficient in the balanced chemical equation: \[ TiCl_4 : { 5.272 \, mol \, (actual) \over 1 \, mol \, (stoich)} = 5.272\nonumber \\[6pt] Mg: {8.23 \, mol \, (actual) \over 2 \, mol \, (stoich)} = 4.12\nonumber \]. Use the given densities to convert from volume to mass. Consider this reaction: 2 C 6 H 14 + 19 O 2 12 CO 2 + 14 H 2 O a. If the actual yield of aspirin is 2, what is the percentage yield? A From the formulas given for the reactants and the products, we see that the chemical equation is balanced as written. Here is a simple and reliable way to identify the limiting reactant in any problem of this sort: Density is the mass per unit volume of a substance. Determine the mass of I2, which could be produced? >u,(8n06SR nCweOSpzUJm/ibR[cQGx ;4j:;('+fB9h6HvJKC)W|C9?6@H&iBWe>4 "t&C"p&N ql;TF/B;I77PE,*4uYV"Kdhguokle'X,V\:P%I*-P9;=&%2
V4c'#MZXh,i&+`0?Id,'MV|!&'. Quantity Excess = Initial Quantity - Consumed Quantity. Use the essential equation sheet as a reference. B To determine which reactant is limiting, we need to know their molar masses, which are calculated from their structural formulas: p-aminobenzoic acid (C7H7NO2), 137.14 g/mol; 2-diethylaminoethanol (C6H15NO), 117.19 g/mol. Twelve eggs is eight more eggs than you need. endobj
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