the mass is 3.25g for 3-nitrophtalic acid, how would you complete the table? ple

Question

the mass is 3.25g for 3-nitrophtalic acid, how would you complete the table? please show all work on how you got rxn weight and mmol. thanks.

2.00 mL mmol Acetic Acid 1.049 g/mL 60.05 (12 pts/1 pt per box) A specific mass of 3-nitrophthalic acid has been preweighed for you by your TA, but it is not the 1.00 g suggested by the textbook. Also, the number of equivalents for reactants and solvents has been modified to optimize the reaction conditions. Using your mass of 3-nitrophthalic acid and the new equivalent values, determine what quanti- ties of reactants and solvents are needed for the specific mass of 3-nitrophthalic acid given to you. These are the quantities you will use for your synthesis of luminol. (Complete the table below immediately after receiving your 3-nitrophthalic acid during 2. lab,) d (g/mL) or Molecular M (mol/L) Reactiorn Reactant or Solvent WeightWeight (g) or mmol Equivalents Volume (mL) (g/mol) 211.13 32.05 150.17 3-Nitrophthalic Acid Hydrazine (8 wt % Aqueous Solution) | 8% by weight Triethylene Glycol 1.125 g/mL Sodium Hydroxide (Aqueous Solution) 3.00 mol/L Sodium Dithionite* g mmol mL mmol mL mmol mmol mmol mL mmol 1.00 1.20 5.00 3.50 4.00 8.00 NA I | 1.00 g/mL 1.00 g/mL NA 174.10 60.05 mL NA Acetic Acid 1.049 g/ml Note: We are using sodium dithionite in our experiment, not sodium dithionite dihydrate. 165

Explanation / Answer

Reagents table

3-nitrophthalic acid = 3.25 g

moles = 3.25 g/211.13 g/mol = 15.4 mmol

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Hydrazine solution

moles = 15.4 mmol x 1.2 = 18.5 mmol

mass = 18.5 mmol x 32.05 g/mol = 0.60 g

mass of 8wt% solution = 0.60 g x 100/8 = 7.5 g

volume of hydrazine solution = 7.5 g/1.0 g/ml = 7.5 ml

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triethylene glycol

moles = 15.4 mmol x 5 = 77.0 mmol

mass = 77 mmol x 150.17 g/mol/1000 = 11.6 g

volume = 11.6 g/1.125 g/ml = 10.3 ml

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sodium hydroxide solution

moles = 15.4 mmol x 3.5 = 53.9 mmol

volume = 53.9 mmol/3 M = 18.0 ml

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Sodium dithionite

moles = 15.4 mmol x 4 = 61.6 mmol

mass = 61.6 mmol x 174.70 g/mol/1000 = 10.8 g

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Acetic acid

moles = 15.4 mmol x 8 = 123.2 mmol

mass acetic acid = 123.2 mmol x 60.05 g/mol/1000 = 7.4 g

Volume = 7.4 g/1.049 g/ml = 7.05 ml

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