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Calculate the moles number of the prepared dye (by finding the limiting reagent). and writing of the formula for the chemical equation

Calculate the moles number of the prepared dye (by finding the limiting reagent). and writing of the formula for the chemical equation

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Calculate the moles number of the prepared dye (by finding the limiting reagent). and writing of the formula for the chemical equation
The calculation: 1- Calculation of the volume of aniline: Aniline -> C6H5NH₂ 93.13 g/mol (0.005 mole) 0.005 mole x 93.13 g/mol = 0.46565 gm 0.46565 gm x 1.02 g ~ = 0.5 ml 2- The calculation of the number of sodium nitrite mole: Sodium nitrite →→ NaNO₂ 69 gm / mol (0.4 gm in 5 ml of water) 0.4 gm ÷ 69 gm /mol = 0.00579 mol 3- The calculation of the number of 2-naphthol grams: 2-naphthol -> C10H8O 144.17 g/mol 0.005 mole x 144.17 g/mol = 0.720 gm 4- The calculation of the moles number of dye: 1 mole of aniline and 1 mole of 2-naphthol → 1 mole of 2-naphthol aniline dye 0.005 mil of aniline 0.005 mole of dye 5- The calculation of the number of dye grams (theoretical yield): Mw of dye 248.28 g/mol x 0.005 mol = 1.241 gm
Transcribed Image Text:The calculation: 1- Calculation of the volume of aniline: Aniline -> C6H5NH₂ 93.13 g/mol (0.005 mole) 0.005 mole x 93.13 g/mol = 0.46565 gm 0.46565 gm x 1.02 g ~ = 0.5 ml 2- The calculation of the number of sodium nitrite mole: Sodium nitrite →→ NaNO₂ 69 gm / mol (0.4 gm in 5 ml of water) 0.4 gm ÷ 69 gm /mol = 0.00579 mol 3- The calculation of the number of 2-naphthol grams: 2-naphthol -> C10H8O 144.17 g/mol 0.005 mole x 144.17 g/mol = 0.720 gm 4- The calculation of the moles number of dye: 1 mole of aniline and 1 mole of 2-naphthol → 1 mole of 2-naphthol aniline dye 0.005 mil of aniline 0.005 mole of dye 5- The calculation of the number of dye grams (theoretical yield): Mw of dye 248.28 g/mol x 0.005 mol = 1.241 gm
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