CHEM 30A Indepdendent Project

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San Jose State University *

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30A

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Chemistry

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Feb 20, 2024

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Independent Project Maili Nguyen Tingle Introductory Chemistry Professor Irina Slobodov December 1, 2023
1 Title: Dilution to customized concentration of Hydrochloric Acid and Iron (III) Chloride Introduction/Objective: The ability to dilute solutions to the needed and appropriate concentrations will be reflected in this lab. Today, we will be working with the following elements HCL (hydrochloric acid) and FeCl 3 (Iron (III) Chloride). The solution in 100 mL should be homogenous and include 0.095 M FeCl 3 and 0.125 M HCL in diluted proportions. Lab Materials: 1. A 100 mL volumetric flask 2. Tablespoon 3. 50 ml graduated cylinder 4. Digital scale 5. 50 mL beaker 6. Protective eyewear 7. Full length pants and full length shirt 8. Isotonic water (H2O) Protocol: 1. The 50 mL graduated cylinder can hold up to 4.2 mL of 3 M HCl 2. Add the 4.2 mL of 3 M HCl into the 100 mL volumetric flask 3. Measure 2.6 g of FeCl 3 6 H 2 O using a tablespoon and digital scale
2 4. Add the measure 2.6 g of FeCl 3 6 H 2 O into the 100 mL volumetric flask with the HCl 5. Add H2O (pure water) into the 100 mL volumetric flask until the mark (100 mL). 6. Swirl the volumetric flask to form a homogeneous solution 7. Record the final concentration Calculations : Required Volume of 3 HCl for dilution: M 1 V 1 = M 2 V 2 3.0 M HCl ∗( V 1 )= 0.125 M 100 mL V 1 = 0.125 M 100 3 = 4.2 mL 4.2 mL of HCl is required for the dilution Required number of moles of solid FeCl3 * 6H2O: 0.095 M FeCl 3 = x 0.100 L = 0.0095 moles FeCl3 0.0095 moles FeCl3 is required Mole to Gram conversion of Solid FeCl3 * 6H2O: 0.0095 FeCl 3 270.38 1 mol FeCl 3 6 H 2 O = 2.6 gFeCl 3 6 H 2 O 2.6 g FeCL3 * 6H2O is required
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