Balancing Equations and Simple Stoichiometry

University:
Felician University
Course:
CHEM-105 | Principles of Chemistry
Academic year:

2024
Views:

148

Pages:

3
Author:

Clint W.

**Stoichiometry Practice Worksheet** **Balancing Equations and Simple Stoichiometry** 5 HW pts! **Balance the following equations:** 1) N₂ + 3F₂ → 2 NF₃ 2) 2 C₆H₁₀ + 17 O₂ → 12 CO₂ + 10 H₂O 3) HBr + KHCO₃ → H₂O + KBr + CO₂ (balanced) 4) 2 GaBr₃ + 3 Na₂SO₃ → Ga₂(SO₃)₃ + 6 NaBr 5) 3 SnO + 2 NF₃ → 3 SnF₂ + N₂O₃ **Solve the following stoichiometry grams-grams problems:** 6) Using the following equation: 2 NaOH + H₂SO₄ → 2 H₂O + Na₂SO₄ How many grams of sodium sulfate will be formed if you start with 200 grams of sodium hydroxide and you have an excess of sulfuric acid? 200g NaOH * (1 mol NaOH / 40.00 g NaOH) * (1 mol Na₂SO₄ / 2 mol NaOH) * (142.04 g Na₂SO₄ / 1 mol Na₂SO₄) = 355 g Na₂SO₄ ≈ 400 g Na₂SO₄ 7) Using the following equation: Pb(SO₄)₂ + 4 LiNO₃ → Pb(NO₃)₄ + 2 Li₂SO₄ How many grams of lithium nitrate will be needed to make 250 grams of lithium sulfate, assuming that you have an adequate amount of lead (IV) sulfate to do the reaction? 250g Li₂SO₄ * (1 mol Li₂SO₄ / 109.9 g Li₂SO₄) * (4 mol LiNO₃ / 2 mol Li₂SO₄) * (68.99 g LiNO₃ / 1 mol LiNO₃) = 313.9 g LiNO₃ ≈ 310 g LiNO₃ Use the following equation to answer questions 8-11: 2 C6H10 + 17 O2 → 12 CO2 + 10 H2O 8) If I do this reaction with 35 grams of C6H10 and 45 grams of oxygen, how many grams of carbon dioxide will be formed? 35g C6H10 x (1 mol C6H10 / 82.14 g C6H10) x (12 mol CO2 / 2 mol C6H10) x (44.01 g CO2 / 1 mol CO2) = 43.6 g CO2 45g O2 x (1 mol O2 / 32 g O2) x (12 mol CO2 / 17 mol O2) x (44.01 g CO2 / 1 mol CO2) = 44.9 g CO2 LR = O2 9) What is the limiting reagent for problem 6? O2 10) How much of the excess reagent is left over after the reaction from problem 6 is finished? 45g O2 x (1 mol O2 / 32 g O2) x (2 mol C6H10 / 17 mol O2) x (82.14 g C6H10 / 1 mol C6H10) = 9.6 g C6H10 35g - 9.6g = 25.4 g excess C6H10 11) If 35 grams of carbon dioxide are actually formed from the reaction in problem 6, what is the percent yield of this reaction? 35g CO2 / 44.9 g CO2 x 100 = 79.5 % **Answer the following stoichiometry-related questions:** 12) Write the balanced equation for the reaction of acetic acid with aluminum hydroxide to form water and aluminum acetate: 3CH₃COOH + Al(OH)₃ → 3H₂O + Al(C₂H₃O₂)₃ 13) Using the equation from problem #12, determine the mass of aluminum acetate that can be made if I do this reaction with 125 grams of acetic acid and 275 grams of aluminum hydroxide. 125g CH₃COOH x (1 mol CH₃COOH / 60.06 g CH₃COOH) = 2.08 mol CH₃COOH 275g Al(OH)₃ x (1 mol Al(OH)₃ / 78.00 g Al(OH)₃) = 3.53 mol Al(OH)₃ 2.08 mol CH₃COOH x (1 mol Al(C₂H₃O₂)₃ / 3 mol CH₃COOH) x (204.13 g Al(C₂H₃O₂)₃ / 1 mol Al(C₂H₃O₂)₃) = 141.5 g Al(C₂H₃O₂)₃ sig figs → 142 g Al(C₂H₃O₂)₃ 14) What is the limiting reagent in problem #13? CH₃COOH (acetic acid) 15) How much of the excess reagent will be left over in problem #13 after the reaction is complete? 125g CH₃COOH x (1 mol CH₃COOH / 60.06 g CH₃COOH) x (1 mol Al(OH)₃ / 3 mol CH₃COOH) x (78.00 g Al(OH)₃ / 1 mol Al(OH)₃) = 54.0 g Al(OH)₃ 275g - 54.0g = 221g excess Al(OH)₃