University:
University of North Carolina at Charlotte
Course:
CHEM 1251 | General Chemistry I
Academic year:

2024
Views:

275

Pages:

13
Author:

Kamila P.

Balanced Chemical Equation: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) Given Information: 11.8 g of NaCl produced 15.3 g HCl 30.6 g NaOH Calculations: Step 1: Calculate moles of HCl and NaOH: 15.3 g HCl * (1 mol HCl / 36.46 g HCl) = 0.4196 mol HCl 30.6 g NaOH * (1 mol NaOH / 40.00 g NaOH) = 0.765 mol NaOH Step 2: Determine the limiting reactant: Based on the balanced equation, 1 mol HCl reacts with 1 mol NaOH. 0.4196 mol HCl is less than 0.765 mol NaOH, so HCl is the limiting reactant. Step 3: Calculate theoretical yield of NaCl: 0.4196 mol HCl * (1 mol NaCl / 1 mol HCl) * (58.44 g NaCl / 1 mol NaCl) = 24.5249 g NaCl Step 4: Calculate percent yield: (11.8 g NaCl / 24.5249 g NaCl) * 100% = 48.12% Additional Calculations: Calculations for H₂SO₄ and NaOH: 2.0 g H₂SO₄ * (1 mol H₂SO₄ / 98.08 g H₂SO₄) = 0.0204 mol H₂SO₄ 0.0204 mol H₂SO₄ * (2 mol NaOH / 1 mol H₂SO₄) = 0.0408 mol NaOH 0.0165 mol NaOH (used in previous calculation) Total NaOH used: 0.0408 mol + 0.0165 mol = 0.0573 mol NaOH 0.0573 mol NaOH * (40.00 g NaOH / 1 mol NaOH) = 2.292 g NaOH Calculations for HCl and NaOH: 0.065 mol NaOH * (1 mol HCl / 1 mol NaOH) = 0.065 mol HCl 0.065 mol HCl * (36.46 g HCl / 1 mol HCl) = 2.375 g HCl Calculations for CO₂: 0.5535 mol CO₂ * (44.01 g CO₂ / 1 mol CO₂) = 24.35 g CO₂ Percent yield of CO₂: (17.2 g CO₂ / 24.35 g CO₂) * 100% = 70.7% 2(CH₃(CH₂)₆CH₃) + 15O₂ → 8CO₂ + 9H₂O 2(CH₃(CH₂)₆CH₃) + 25O₂ → 16CO₂ + 18H₂O 4.69 g CH₃(CH₂)₆CH₃ × 1 mol CH₃(CH₂)₆CH₃ / 144.2249 g CH₃(CH₂)₆CH₃ = 0.0402777 mol CH₃(CH₂)₆CH₃ 8.03 g O₂ × 1 mol O₂ / 32 g O₂ = 0.25 mol O₂ 0.0402777 mol CH₃(CH₂)₆CH₃ × 18 mol H₂O / 2 mol CH₃(CH₂)₆CH₃ = 0.3624 mol H₂O 0.25 mol O₂ × 18 mol H₂O / 25 mol O₂ = 0.18 mol H₂O (THEORETICAL YIELD) 0.18 mol H₂O × 18.01528 g H₂O / 1 mol H₂O = 3.24 g H₂O 9 H₂O EXCESS: 0.16 mol H₂O / 18 mol H₂O = 0.002 mol CH₃(CH₂)₆CH₃ 0.0402777 mol CH₃(CH₂)₆CH₃ - 0.002 mol CH₃(CH₂)₆CH₃ = 0.0202777 mol CH₃(CH₂)₆CH₃ HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) 7.29 g HCl × 1 mol HCl / 36.46 g HCl = 0.19996 mol HCl 2.89 g NaOH × 1 mol NaOH / 39.997 g NaOH = 0.075173 mol NaOH 0.19996 mol HCl × 1 mol NaCl / 1 mol HCl = 0.19996 mol NaCl 0.075173 mol NaOH × 1 mol NaCl / 1 mol NaOH = 0.075173 mol NaCl 0.075173 mol NaOH × 58.359 g NaCl / 1 mol NaCl = 4.39 g NaCl 2(CH₃CH₃)(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g) 2.45 g CH₃CH₃ × 1 mol CH₃CH₃ / 30.068 g CH₃CH₃ = 0.079819 mol CH₃CH₃ 17.89 g O₂ × 1 mol O₂ / 32 g O₂ = 0.55625 mol O₂ 0.079819 mol ethene × 4 mol CO₂ / 2 mol ethene = 0.159638 mol CO₂ 0.55625 mol O₂ × 4 mol CO₂ / 7 mol O₂ = 0.317857 mol CO₂ Hexane 2(CH₃(CH₂)₄CH₃) + 19O₂(g) → 12CO₂(g) + 14H₂O(g) 7.76 g hexane × 1 mol hexane / 86.172 g hexane = 0.09005036382 mol hexane 16.6 g O₂ × 1 mol O₂ / 32 g O₂ = 0.5 mol O₂ 0.09005036382 mol hexane × 14 mol H₂O / 2 mol hexane = 0.63035 mol H₂O 0.5 mol O₂ × 14 mol H₂O / 19 mol O₂ = 0.368421 mol H₂O (THEORETICAL YIELD) 0.368421 mol H₂O × 18.01528 g H₂O / 1 mol H₂O = 6.6 g H₂O If 39 g of carbon dioxide is produced from 6.0 g hexane and 6.4 g oxygen, what is the percent yield of CO₂? 6.4 g oxygen × 1 mol O₂ / 32 g O₂ = 0.2 mol O₂ 0.2 mol O₂ × 12 mol CO₂ / 19 mol O₂ = 0.126316 mol CO₂ 39 g CO₂ / 0.126316 mol CO₂ = 308.57 g/mol CO₂ 308.57 g/mol CO₂ × 100% / 44.01 g/mol CO₂ = 701%