Chapter 3: Stoichiometry 3: Stoichiometry 5: Thermochemistry 8: Covalent Bonding and Molecular Structure 15: Chemical Equilibrium 16: Acids and Bases 3.2 Stoichiometry and Compound Formulas 3.1 The Mole and Molar Mass 3.2 Stoichiometry and Compound Formulas 3.3 Stoichiometry and Chemical Reactions 3.4 Stoichiometry and Limiting Reactants 3.5 Chemical Analysis Chapter Summary Chapter Summary Assignment Reference Tools Periodic Table Molarity Calculator Molar Mass Calculator Unit
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In the Stoichiometry Challenge Lab we compared the theoretical results of the reaction between sodium carbonate (Na2CO3) and sulfuric acid (H2SO4) with the actual data we found. I hypothesised that If the mole ratio between Na2SO4 and H2SO4 is 1:1 then when I react 0.5 grams of Na2SO4 (reactant with H2SO4) I should get 0.669 grams of Na2SO4. The actual reaction between .05 grams of Na2CO3 and 5 mL of of H2SO4 produced 0.79g of Na2SO4. When I were testing the reaction‚ I measured out the reactants
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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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Hugh Kim Lab Report: Stoichiometry Lab 1. Prelab Part1. 1) Create no waste = The principle that encourages chemists to not create waste at the first place rather than cleaning it up afterwards effectively shifts the chemistry more environmentally conscious‚ as creating no waste would make the experiment efficient; the reactants will be reduced to only the essential ones and the product will be maximized‚ a change that would make the experiment economic. Also‚ if chemists aim to
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Basic Stoichiometry PhET Lab rvsd 2/2011 Let’s make some sandviches! _ Introduction: When we bake/cook something‚ we use a specific amount of each ingredient. Imagine if you made a batch of cookies and used way too many eggs‚ or not enough sugar. YUCK! In chemistry‚ reactions proceed with very specific recipes. The study of these recipes is stoichiometry. When the reactants are present in the correct amounts‚ the reaction will produce products. What happens if there are more or less of
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ordinarily able to measure only the total pressure of a gaseous mixture‚ so if two or more gaseous products are present in the equilibrium mixture‚ the partial pressure of one may need to be inferred from that of the other‚ taking into account the stoichiometry of the
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The calculations completed for this 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
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Title: Stoichiometry Reaction Objectives: 1. To decompose sodium hydrogen carbonate (sodium bicarbonate) by heating. 2. To accurately measure the degree of completion of the reaction by analysing the solid sodium carbonate product. 3. To calculate amount of product with given amount of reactant. 4. To determine amount of heat release in the reaction. Results: Part 1: Thermal Decomposition of NaHCO3 Materials Mass (g) Clean and dry test tube 15.1632 Clean test tube + NaHCO3 17.1647
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Na2CO3(s)+CO2(g)+H2O(g). The was the only equation that matched up exactly with my data in terms of percentage. To start off with‚ when I balanced out the equation‚ I got 2 NaHCO3→ 1 Na2CO3(s)+ 1 CO2(g)+ 1 H2O(g). Therefore when I set up my stoichiometry problem I got 3.2 grams NaHCO3 over 1 x 1 mol NaHCO3 over 84.007g NaHCO3 x 1mol Na2CO3 over 2 mol NaHCO3 x 105.987g Na2CO3 x 1 mol Na2CO3. Hence‚ I multiplied 3.2 x 1 x 1 x 105.987 and got 339.1584. Afterwards‚ I divided 339.1584 by 84.007 and
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Lab 3: Stoichiometry of a Precipitation Reaction NOTE: All photos should be taken so the reading on the electronic balance is readable. Photo 1: filter paper being weighed on electronic balance‚ along with your student information card. Photo 2: beaker with precipitate slurry in it (after step 5) ‚ along with your student information card Photo 3: dried precipitate/filter paper being weighed on electronic balance‚ along with your student information card Additional Question Guidelines:
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