Chemistry 1: PROBLEM SET SY 2012/2013 CLASS #: _______________ NAME_________________________________________ SECTION: ______________________ Stoichiometry II - Mole Calculations/ Limiting and Excess Reagent – Lecture Notes 1. Given the balanced equation N2(g) + 3H2(g) 2NH3(g) How many moles of ammonia are produced when 0.60 mol of nitrogen reacts with hydrogen? 2. Given the equation: SiO2 + HF SiF4 + H2O a. Calculate the number of moles HF that would completely react with 2.5 moles of
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Stoichiometry February 28th‚ 2013 Abstract: The reactions of the Sodium Hydroxide and two acids‚ Hydrochloric Acid and Sulfuric Acid were performed. The heat given off by these two reactions was used to determine the stoichiometric ratio and the limiting reactants in each experiment. Introduction: Coefficients in a balanced equations show how many moles of each reactant is needed to react with each other and how many moles of each product that will be formed. Stoichiometry allows us to
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purposefully combining the elements‚ instead of having it naturally do so‚ made a significantly fully more bright color. In Theophilus’s “recipe” for vermillion‚ it is stated that there should be far more sulfur than a stiochiometric reaction should require. Stoichiometry
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experiment include determining the amount of Na2CO3 needed to do a full reaction. This was calculated through stoichiometry calculations: Molar mass was first calculated for CaCl2*2H2O Ca = 40.078g Cl2 = 35.453g*2 = 70.906g 2H2 = 1.00794g*4 = 4.03176g 2O = 15.9994g*2 = 31.9988g 40.078g + 70.906g + 4.03176g + 31.9988g = 147.01456g or 147.0 g CaCl2 1g CaCl2 * 2H2O x (1 mol CaCl2 *2H2O/147g CaCl2 *2H2O) = 0.0068 mol of CaCl2*2H2O Molar mass was then calculated for Na2CO3: Na2 = 22.9898g*2
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Stoichiometry Lab Name Questions A. From your balanced equation‚ what is the theoretical yield of your product? Theoretical yield of the CaCO3 is expected to be .69g. B. According to your data table‚ what is the actual yield of the product? The mass of the filter paper was 1.1g‚ and the total mass of the filter paper when dried with the CaCO3 was 1.8 total. Thus the actual yield of the product was .70g. C. What is the percent yield? Percentage yield is actual yield over the theoretical
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moles-stoichiometry-practice-problems Now you’re ready to use what you know about conversion factors to solve some stoichiometric problems in chemistry. Almost all stoichiometric problems can be solved in just four simple steps: 1.Balance the equation. 2.Convert units of a given substance to moles. 3.Using the mole ratio‚ calculate the moles of substance yielded by the reaction. 4.Convert moles of wanted substance to desired units. These "simple" steps probably look complicated at first
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STOICHIOMETRY OF GASOLINE. AN INTRODUCTION The internal combustion engines burn fuel to create kinetic energy. The burning of fuel is basically the reaction of fuel with oxygen in the air to form water and Carbon dioxide as the major end product . The amount of oxygen present in the cylinder is the limiting factor for the amount of fuel that can be burnt that is to say it determines the level of burning in our combustion engine. If there’s too much fuel present‚ not all fuel will be burnt and un-burnt
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Stoichiometry and Limiting Reagents Theodore A. Bieniosek I. Purpose and Theory The purpose of the experiment is to study and apply the processes of stoichiometric calculation on a controlled chemical reaction. We will be adding variable amounts of reactants in a chemical reaction in order to demonstrate the effect of limiting reagents. Based on the volumes of the reactants‚ and their respective molarities‚ we can calculate the theoretical yield of the reaction and compare it to the
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Chapter 4 — Intro—1 1 CHAPTER 3 Topic Scopes: Stoichiometry and Solution Concentration • Molarity‚ molality‚ parts per million & percentage (w/w‚ w/v and v/v) • Stoichiometry calculation • Limiting reactant • Theoretical yield‚ actual yield and percentage yield 1 2 Mole Concept No. of Moles = Molarity (M) • Molarity (molar concentration) is the number of moles of a solute that is contained in 1 liter of solution Mass (g) molar mass (g/mol) No. of Moles = Molarity (mol/L) volume (L) Molarity
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Chem 121L Part I: Introduction Stoichiometry is the study of the quantitative‚ or measurable‚ relationships that exist in chemical formulas and also chemical reactions. In this experiment hydrogen gas will be produced from the reaction of a known mass of magnesium metal with an excess of hydrochloric acid. The theoretical number of moles of hydrogen gas may be calculated using stoichiometry and the balanced chemical equation. The theoretical volume of hydrogen gas may then be determined from
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