Unit 3- Lab Activity -Stoichiometry and Gas Laws applications

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Dec 6, 2023

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Unit 3: Lab activity NAME: Stoichiometry – Chapter 10 (Limiting reactants and percent yields) Stoichiometry – Chapter 14 Gas Law 1. According to the following reaction. a ) How many grams of CO 2 and H 2 O can be produced from a reacting 10.1 g of O 2 and 21.4 g of C 4 H 10 ? (Hint: Balance the reactions & find the limiting reactant) 13 O 2 + 2 C 4 H 10 → 8 CO 2 + 10 H 2 O = 0.1942 mol CO 2 = 1.473 mol CO 2 Mass CO 2 : 0.1942 x 44.01 = 8.55 g CO 2 Mass H 2 O: = 4.38 g H 2 O 10.1 g O 2 1 mol O 2 8 mol CO 2 32.00 g O 2 13 mol O 2 21.4 g C 4 H 10 1 mol C 4 H 10 8 mol CO 2 58.12 C 4 H10 2 mol C 4 H 10 10.1 g O 2 1 mol O 2 10 mol H 2 O 18.02 g H 2 O 32.00 g O 2 13 mol O 2 1 mol H 2 O
b) How many grams will be left over that did not react? = 2.82 g C 4 H 10 reacted 21.4 g – 2.82 g = 18.58 g C 4 H 10 left over that did not react. 2. Ammonia can be made by reaction of water with magnesium nitride as shown below: Mg 3 N 2 ( s ) + 6 H 2 O ( l ) → 3 Mg(OH) 2 ( s ) + 2 NH 3 ( g ) a) How grams of ammonia and magnesium hydroxide can be formed when 4.66 g Mg 3 N 2 and 23.0 g H 2 O are reacted? (Hint: Balance the reactions & find the limiting reactant) = 0.139 mol Mg(OH) 2 = 0.638 mol Mg(OH) 2 10.1 g O 2 1 mol O 2 2 mol C 4 H 10 58.12 C 4 H 10 32.00 g O 2 13 mol O 2 1 mol C 4 H 10 4.66 g Mg 3 N 2 1 mol Mg 3 N 2 3 mol Mg(OH) 2 100.93 g Mg 3 N 2 1 mol Mg 3 N 2 23.0 g H 2 O 1 mol H 2 O 3 mol Mg(OH) 2 18.015 g H 2 O 6 mol H 2 O
Mass Mg(OH) 2 : 0.139 x 58.32 = 8.106 g Mg(OH) 2 Mass NH 3 : = 1.573 g NH 3 b) How many grams will be left over that did not react? = 4.99 g H 2 O reacted 23.0 g – 4.99 g = 18.01 g H 2 O left over that did not react. 3. When 50.0 g of Fe 2 O 3 react with excess Al according to the following reaction. Fe 2 O 3 (s) + 2 Al(s) → Al 2 O 3 (s) + 2 Fe(s) a) Determine the mass (g) of Fe produced? = 34.97 g Fe 4.66 g Mg 3 N 2 1 mol Mg 3 N 2 2 mol NH 3 17.03 g NH 3 100.93 g Mg 3 N 2 1 mol Mg 3 N 2 1 mol NH 3 4.66 g Mg 3 N 2 1 mol Mg 3 N 2 6 mol H 2 O 18.015 g H 2 O 100.93 g Mg 3 N 2 1 mol Mg 3 N 2 1 mol H 2 O 50.0 g Fe 2 O 3 1 mol Fe 2 O 3 2 Fe 55.845 g Fe 159.688 g Fe 2 O 3 1 mol Fe 2 O 3 1 mol Fe
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b) Determine the percent yield of the reaction produces 30.6 g of Fe. % yield = × 100 = 87.5% 4. According to the following reaction. N 2 O 4 (l) + 2 N 2 H 4 (l) → 3 N 2 (g) + 4 H 2 O(g) a) Determine the limiting reactant (LR) and the mass (g) of nitrogen that can be formed from 40.0 g N 2 O 4 and 50.0 g N 2 H 4 . = 1.304 mol N 2 = 2.34 mol N 2 N 2 O 4 is the limiting reactant. Mass of N 2 : 1.304 x 28.013 = 36.529 g N 2 . 40.0 g N 2 O 4 1 mol N 2 O 4 3 mol N 2 92.011 g N 2 O 4 1 mol N 2 O 4 50.0 g N 2 H 4 1 mol N 2 H 4 3 mol N 2 32.045 g N 2 H 4 2 mol N 2 H 4
5. Determine the theoretical yield (grams) of HCl if 120.0 g of BCl 3 and 90.5 g of H 2 O are reacted according to the following balanced reaction. BCl 3 (g) + 3 H 2 O(l) → H 3 BO 3 (s) + 3 HCl(g) Determine limiting reactant (LR) and the mass (g) of HCl produced. = 3.07 mol HCl = 5.02 mol HCl BCl 3 is the limiting reactant. Mass of HCl: 3.07 x 36.46 = 111.93 g HCl 6. When 10.0 g C4H10 reacted with 30.0 g of O2 according to the following reaction. a) Balance the following equation. 2 C 4 H 10 + 13 O 2 → 10 H 2 O + 8 CO 2 ∆H = - 5750 KJ b) Determine the limiting reactant. = 0.688 mol CO 2 120.0 g BCl 3 1 mol BCl 3 3 mol HCl 117.17 g BCl 3 1 mol BCl 3 90.5 g H 2 O 1 mol H 2 O 3 mol HCl 18.015 g H 2 O 3 mol H 2 O 10.0 g C 4 H 10 1 mol C 4 H 10 8 mol CO 2 58.12 g C 4 H 10 2 mol C 4 H 10
= 0.577 mol CO 2 O 2 is the limiting reactant. c) Determine the mass grams of CO2 and H2O produced from the reaction. Mass of CO 2 : 0.577 x 44.01 = 25.39 g CO 2 Mass of H 2 O: = 12.99 g H 2 O d) Determine the energy produced from the reaction. = 414.66 KJ 7. When 680.0 g C8H18 is reacted with 280.0 g O2 gas according to the following reaction. a) Balance the following equation. 30.0 g O 2 1 mol O 2 8 mol CO 2 32.00 g O 2 13 mol O 2 30.0 g O 2 1 mol O 2 10 mol H 2 O 18.015 g H 2 O 32.00 g O 2 13 mol O 2 1 mol H 2 O 30.0 g O 2 1 mol O 2 5750 KJ 32.00 mol O 2 13 mol O 2
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2 C 8 H 18 + 25 O 2 → 18 H 2 O + 16 CO 2 ∆H = - 10900 KJ b) Determine the limiting reactant. = 53.58 mol H 2 O = 6.3 mol H 2 O O 2 is limiting reactant. c) Determine the mass, grams of CO2 and H2O produced from the reaction. Mass of H 2 O: 6.3 x 18.015 = 113.49 g H 2 O Mass of CO 2 : = 246.46 g CO 2 d) Determine the energy produced from the reaction. = 3815 KJ 680.0 g C 8 H 18 1 mol C 8 H 18 18 mol H 2 O 114.23 g C 8 H 18 2 mol C 8 H 18 280.0 g O 2 1 mol O 2 18 mol H 2 O 32.00 mol O 2 25 mol O 2 280.0 g O 2 1 mol O 2 16 mol CO 2 44.01 g CO 2 32.00 mol O 2 25 mol O 2 1 mol CO 2 280.0 g O 2 1 mol O 2 10900 KJ 32.00 mol O 2 25 mol O 2
8. Sodium azide (NaN 3 ) when ignited form sodium metal and nitrogen gas according to the following equation: NaN 3(s) → Na (s) + N 2(g) (a) Balance the chemical equation: 2 NaN 3 → 2 Na + 3 N 2 (b) Identify the type of reaction: decomposition. (c) How many grams of NaN 3 must be used to produce 85.5 L of N 2 gas at 28 o C and 95.0 psi of pressure? 28 o C = 301 K 95 psi = 6.46437 atm PV = nRT n = 𝑃? = 6.46437 × 85.5 = 22.366 mol N 2 𝑅? 0.0821 ×301 = 14.91 mol NaN 3 Mass of NaN 3 : 14.91 x 65.01 = 696.299 g NaN 3 22.366 mol N 2 3 mol NaN 3 2 mol N 2
9. In the lab, students generated and collected hydrogen gas from the reaction of Mg with hydrochloric acid. a) Write a balanced chemical equation of the reaction. Mg(s) + 2 HCl → MgCl 2 + H 2 (g) b) What volume (L) of hydrogen gas at STP were generated from 42.9 g of Mg metal? Use STP 1 mol = 22.4 L = 39.54L H 2 produced. 10. Consider the following equation: C 6 H 14 + O 2 → CO 2 + H 2 O a) Balance the chemical equation. 2 C 6 H 14 + 19 O 2 → 12 CO 2 + 14 H 2 O b) How many liters of oxygen gas are required to react with 9.45 L C 6 H 14 if both gases are at STP? = 89.775 L O 2 are required. 42.9 g Mg 1 mol Mg 1 mol H 2 22.4 L 24.305 g Mg 1 mol Mg 1 mol H 2 9.45 L C 6 H 14 1 mol C 6 H 14 19 mol O 2 22.4 L 22.4 L C 6 H 14 2 mol C 6 H 14 1 mol O 2
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c) How many grams of CO 2 will be produced from 12.0 L C 6 H 14 at 25.0 0 C and 1.84 atm? 25 0 C = 298 K PV = nRT => n = 𝑃? = 1.84×12 = 0.904 mol C 6 H 14 𝑅? 0.082×298 = 5.424 mol CO 2 Mass of CO 2 : 5.424 x 44.01 = 238.71 g CO 2 d) How many grams of water vapor can be produced when 1.25 L C 4 H 14 at 15.0 0 C and 1.15 atm are reacted with 2.00 L O 2 at ─10.0 0 C and 20.2 psi? 1 atm = 14.7 psi 20.2 psi = 1.374 atm -10 0 C = 263 K 15 0 C = 288 K 2 C 4 H 14 + 15 O 2 → 8 CO 2 + 14 H 2 O nO 2 = 𝑃? = 1.374×2 = 0.127 mol O 2 0.904 mol C 6 H 14 12 mol CO 2 2 mol C 6 H 14
𝑅? 0.082×263 nC 4 H 14 = 1.15×1.25 = 0.061 mol C 4 H 14 0.082×288 = 0.1185 mol H 2 O Mass of H 2 O: 0.1185 x 18.015 = 2.135 g H 2 O 0.127 mol O 2 14 mol H 2 O 15 mol O 2

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