Stoichiometry Practice Answers 4 Questions | Felician University

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

2020
Views:

133

Pages:

1
Author:

Clint W.

Stoichiometry Practice 1. Tin(II) fluoride, stannous fluoride, is used in some home dental treatment products. It is made by reacting tin with hydrogen fluoride according to the equation below. Sn(s) + 2HF(g) → SnF₂(s) + H₂(g) If 126 g of tin are used, how many moles of tin(II) fluoride, SnF₂, can be produced? 126 g Sn × 1 mol Sn / 118.79 g Sn × 1 mol SnF₂ / 1 mol Sn = 1.06 mol SnF₂ 2. Magnesium burns in oxygen to produce magnesium oxide. a. Write a balanced equation. 2 Mg(s) + O₂(g) → 2 MgO(s) b. How many moles of oxygen gas are needed to burn 2.5 grams of magnesium? 2.5 g Mg × 1 mol Mg / 24.31 g Mg × 1 mol O₂ / 2 mol Mg = 0.051 mol O₂ 3. Laughing gas (nitrous oxide, N₂O) is sometimes used as an anesthetic in dental work. It is produced when ammonium nitrate is decomposed. Another product is water. a. Write a balanced equation. NH₄NO₃(s) → N₂O(g) + 2H₂O(l) b. How many grams of ammonium nitrate is required to produce 33.0 g of nitrous oxide? 33.0 g N₂O × 1 mol N₂O / 44.02 g N₂O × 1 mol NH₄NO₃ / 1 mol N₂O × 80.05 g NH₄NO₃ / 1 mol NH₄NO₃ = 60.0 g NH₄NO₃ c. How many grams of water are produced in this reaction? 33.0 g N₂O × 1 mol N₂O / 44.02 g N₂O × 2 mol H₂O / 1 mol N₂O × 18.02 g H₂O / 1 mol H₂O = 27.0 g H₂O 4. When sodium reacts with water, hydrogen gas and sodium hydroxide are produced. a. Write a balanced equation. 2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g) b. Determine the mass of sodium hydroxide when 0.25 g of sodium reacts with water. 0.25 g Na × 1 mol Na / 22.99 g Na × 2 mol NaOH / 2 mol Na × 39.998 g NaOH / 1 mol NaOH = 0.44 g NaOH c. Calculate the number of water molecules required to produce 10.0 L of hydrogen gas at STP. 10.0 L H₂ × 1 mol H₂ / 22.4 L H₂ × 2 mol H₂O / 1 mol H₂ × 6.022 × 10^23 molecules H₂O / 1 mol H₂O = 5.38 × 10^23 molecules of H₂O d. Determine the mass of the water molecules produced from 3.27 × 10^23 atoms of sodium that react with water. 3.27 × 10^23 atoms Na × 1 mol Na / 6.022 × 10^23 atoms Na × 2 mol H₂O / 2 mol Na × 18.016 g H₂O / 1 mol H₂O = 0.978 g H₂O