Question3: Experiment 3 The unknown acid sample was 1 • Monoprotic Acid Trails Initial NaOH solution (mL) final NaOH solution (mL) The volume of NaOH to titrate the acid (mL) Amount of Unknown Acid sample 1 (g) The moles of the Unknown Acid (mol) Molar mass of the Unknown Acid (g/mol) A 3.38 28.31 24.93 0.150 0.0026 57.69 B 0.18 29.32 29.14 0.175 0.0029
Table 1 Results DDA Concentration Initial Mass(g) Time Interval Recovered Mass Cumulative Mass (g) Cumulative Recovery (%) Ln[(Rinf -R)/ Rinf] R=Rinf(1-e-kt) (M) (g) 10^(-5) 160 0 0
In the lab, the theoretical yield of CO2 was calculated by using the mass of the Alka-Seltzer (in Part A) and the mass of the anti-acid (Part B) were multiplied by the mole ratios that were involved in the reaction equation for each reaction and they were also divided by the molar mass of each substance. From the balanced chemical equations that were included in the calculation portion, the limiting reactant in part A was determined to be citric acid. The limiting reactant in part B was determined to be HCl because in the reaction, the HCl would ran out before the calcium carbonate. The stoichiometry would not have been different when converting between the different states of matter because the state is a measure of kinetic energy, and not
Stoichiometry of a Double Displacement Reaction The objective of this lab is to find the percent yield of a product of a double displacement reaction. Procedure: Refer to handout entitled “Stoichiometry of a Double Displacement Reaction” Materials: Refer to handout entitled “Stoichiometry of a Double Displacement Reaction” Data & Observations: Data Table Calculated Molar Mass of CuSO4•5H2O 249.677 g Calculated Molar Mass of CuO 79.545 g Starting mass of CuSO4•5H2O 2.050 g Mass of 100-ml beaker and filter paper 52.600 g Mass of 100-ml beaker, filter paper, and CuO precipitate 53.450 g Calculations:
The lab started off by measuring critical materials for the lab: the mass of an an empty 100 mL beaker, mass of beaker and copper chloride together(52.30 g), and the mass of three iron nails(2.73 g). The goal of this experiment is to determine the number of moles of copper and iron that would be produced in the reaction of iron and copper(II) chloride, the ratio of moles of iron to moles of copper, and the percent yield of copper produced. 2.00 grams of copper(II) chloride was added in the beaker to mix with 15 mL of distilled water. Then, three dry nails are placed in the copper(II) chloride solution for approximately 25 minutes. The three nails have to be scraped clean by sandpaper to make the surface of the nail shiny; if the nails are not clean, then some unknown substances might accidentally mix into the reaction and cause variations of the result.
Cassie Droelle 1510-01501 Chemical Equation 2 Visualizing Stoichiometry – Inquiry Lab Cassie Droelle 1510-01501 4-18-16 Introduction The purpose of this lab is to determine the complete balanced equation of a metathesis reaction with an unknown metal ion by first experimentally identifying the metal ion and then using stoichiometry to determine the rest of the equation. Stoichiometry is “the process by which quantities in a chemical reaction are compared” and is based on the Law of Conservation of Mass (1). The Law of Conservation of Matter “dictates that the mass of the reactants and mass of the products must be equal” (1).
Lab 27. Stoichiometry and Chemical Reactions Report In our lab we were asked Which Balanced Chemical Equation Best Represents the Thermal Decomposition of Sodium Bicarbonate. Sodium Bicarbonate is a chemical compound with the formula NaHCO3, also known as baking soda. In the process to answer our guiding question we have to determine how atoms are rearranged during a chemical reaction.
Stoichiometry is a method used in chemistry that involves using relationships between reactants and products in a chemical reaction, to determine a desired quantitative data. The purpose of the lab was to devise a method to determine the percent composition of NaHCO3 in an unknown mixture of compounds NaHCO3 and Na2CO. Heating the mixture of these two compounds will cause a decomposition reaction. Solid NaHCO3 chemically decomposes into gaseous carbon dioxide and water, via the following reaction: 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g). The decomposition reaction was performed in a crucible and heated with a Bunsen burner.
a. At 60 mm Hg, the data values in the table are not consistent with the data in the plots; for example, total mm Hg O2 bound is 30 with oxygen for Bonnie and 2 for Clyde. On the conventional plot, the 60 mm Hg is depicted at 50 with oxygen for Bonnie and about 9 for Clyde. The table says Bonnie should be shown at 6 instead of 19 at this concentration on the seating plot; also, the table says Clyde should be shown at 0 additional amount bound instead of 10 on the seating plot. b.
When the percents were ordered from least to greatest, the molarity of each solution can be deduced, as those with the least change had a molarity closest to zero and those with the most change had a molarity that was the highest. This is due to the fact that if they had 0.0M of solute, they are isotonic and should have no net diffusion of water in a single direction. If they have higher concentrations of sucrose, the water potentials are no longer equal and to decrease entropy of the system, water must diffuse to achieve equilibrium. In this way, it was deduced that Solution B is 0.0M, Solution D is 0.2M, Solution F is 0.4M, Solution A is 0.6M, Solution E is 0.8M, and Solution C is 1.0M. The graph seconded this data by portraying an almost constant
In order to create 5 grams of MgSO4 from MgO (Magnesium Oxide) and H2SO4 (Sulfuric acid), we needed to create a balanced equation to find the amount of other chemicals we would need. The balanced equation was MgO + H2SO4 --> MgSO4 + H2O. After creating a balanced equation, we found the amounts of MgO and H2SO4 using stoichiometry. The amount of Magnesium Oxide was 1.674 grams and the amount of Sulfuric acid was 6.923 milliliters. In order to create as close to 5 grams of MgSO4 as possible, we decided to ignore sig-figs, and go to the hundreds place for the sake of exactness (getting an A).
Verna Wang Hannah Palmer CHEM 101-069 Lab 11-19-16 Stoichiometry and Limiting Reagents Lab Report Purpose: We are using the reaction of sodium hydroxide and calcium chloride to illustrate stoichiometry by demonstrating proportions needed to cause a reaction to take place. Background: Just like a recipe would call for a specific amount of one ingredient to a specific amount of another, stoichiometry is the same exact method for calculating moles in a chemical reaction. Sometimes, we may not have enough of or too much of one ingredient , which would be defined as limiting and excess reagent, respectively.
By finding where the first K is equal to 1 and seeing what the Q value is at this given point, K can be solved. As said above when H equals 1 the Q value is the same as the K value. The K value was calculated to be 4.7. In the second method for solving for K, use the values calculated for the variables Q, K, and n and plug into Equation 1. By doing this the calculated value was 4.645.
IV. Data and observations Mass of beaker (g) 174.01 Mass of beaker + NaOH pellets (g) 174.54 Mass of NaOH pellets 0.53 TRIAL 1 TRIAL 2 Mass of potassium acid phtalate (KHP) (g) 0.15 0.15 final buret reading (ml) 30.75
Practical I: Acid-base equilibrium & pH of solutions Aims/Objectives: 1. To determine the pH range where the indicator changes colour. 2. To identify the suitable indicators for different titrations. 3.
