Stoichiometry 93 in Visual C#

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Stoichiometry 93
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Answer: 1 Convert to moles: 1 mol P O 2 5 = 0880 mol P2O5 2 5 1 mol H O 2 (500 g H 2O) = 278 mol H 2O 180 g H 2O
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(125 g P O ) 142 g P O
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2 Find the limiting reactant 0880 mol = 0880 P2O5 1 mol 278 mol = 0927 H 2O 3 mol The 1 mol and the 3 mol come from the balanced chemical equation The 0880 is smaller, so this is the LR 3 Finish using the number of moles of the LR 2 mol H PO 980 g H PO 3 4 3 4 = 172 g 1 mol P2O5 1 mol H3PO 4
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(0880 mol P O )
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Percent Yield
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In the preceding problems, the amount of product calculated based on the limitingreactant concept is the maximum amount of product that could be formed from the given amount of reactants This maximum amount of product formed is called the theoretical yield However, rarely is the amount that is actually formed (the actual yield) the same as the theoretical yield Normally it is less There are many reasons for this, but the principal reason is that most reactions do not go to completion; they establish an equilibrium system (see the Equilibrium chapter for a discussion of chemical equilibrium) For whatever reason, not as much as expected is formed The efficiency of the reaction can be judged by calculating the percent yield The percent yield (% yield) is the actual yield divided by the theoretical yield, and the result is multiplied by 100% to generate percentage: % yield = actual yield 100% theoretical yield
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Consider the problem in which it was calculated that 608 g NH3 could be formed Suppose that reaction was carried out, and only 523 g NH3 was formed What is the percent yield % yield = 523 g 100% = 860% 608 g
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Let s consider another percent yield problem in which a 250-g sample of calcium oxide is heated with excess hydrogen chloride to produce water and 375 g of calcium chloride What is the percent yield of calcium chloride CaO(s) + 2HCl(g) CaCl2(aq) + H2O(l) Answer: 1 mol CaO 1 mol CaCl2 110984 g CaCl2 (250 g CaO) = 49478 g CaCl2 56077 g CaO 1 mol CaO 1 mol CaCl2
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94 U STEP 4 Review the Knowledge You Need to Score High
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The theoretical yield is 495 g 375 g CaCl2 100% = 758% 49478 g CaCl2 Note: All the units except % must cancel This includes canceling g CaCl2 with g CaCl2, not simply g
Molarity and Solution Calculations
We discuss solutions further in the chapter on solutions and colligative properties, but solution stoichiometry is so common on the AP exam that we will discuss it here briefly also Solutions are homogeneous mixtures composed of a solute (substance present in smaller amount) and a solvent (substance present in larger amount) If sodium chloride is dissolved in water, the NaCl is the solute and the water the solvent One important aspect of solutions is their concentration, the amount of solute dissolved in the solvent In the chapter on solutions and colligative properties we will cover several concentration units, but for the purpose of stoichiometry, the only concentration unit we will use at this time is molarity Molarity (M) is defined as the moles of solute per liter of solution: M = mol solute/L solution Let s start with a simple example of calculating molarity A solution of NaCl contains 3912 g of this compound in 1000 mL of solution Calculate the molarity of NaCl Answer: 1 mol NaCl (3912 g NaCl) 5845 g NaCl = 6693 M NaCl 1 L (1000 mL) 1000 mL Knowing the volume of the solution and the molarity allows you to calculate the moles or grams of solute present Next, let s see how we can use molarity to calculate moles How many moles of ammonium ions are in 0100 L of a 020 M ammonium sulfate solution Answer: 020 mol (NH 4 )2 SO4 2 mol NH 4 + + (0100 L) = 0040 mol NH 4 L 1 mol (NH 4 )2 SO4 Stoichiometry problems (including limiting-reactant problems) involving solutions can be worked in the same fashion as before, except that the volume and molarity of the solution must first be converted to moles If 3500 mL of a 01500 M KOH solution is required to titrate 4000 mL of a phosphoric acid solution, what is the concentration of the acid The reaction is: 2KOH (aq) + H3PO4 (aq) K2HPO4 (aq) + 2H2O (l) Answer: (3500 mL ) (01500 mol KOH)(1 mol H3PO4 ) (1000 mL )( 2 mol KOH) = 006562 M H3PO4 (1 L ) 4000 mL ) ( (1000 mL )
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