In this lab I used: 3 pentanol Alcohol as my c5 alcohol which has a molcular wei

Question

In this lab I used:

3 pentanol Alcohol as my c5 alcohol which has a molcular weight of 88.148 g/mol

the mass of my c5 3- pentanol alcohol that we used in the experiment was 4.108 grams

I also used 7.0 mL of glacial acetic acid. The molar mass of glacial acetic acid is 60.05 g/mol

Can someone solve these clearly for me and tell me what the limiting reagent is? thank you

Part Two: Stoichiometric Calulations a) Report the number of moles of C-4/C-5 alcohol used. Show your work (incl. cancellatione ANSWER moles b) Report the number of grams of glacial acetic acid used. Show your work/units below. (1 pt) ANSWER c) Report the number of moles of glacial acetic acid used. Show your work/units below. (1 pt) ANSWER moles d) The limiting reagent is: (1 pt)

Explanation / Answer

A)

mol of pentanol = mass/MW = 4.108/88.148 = 0.04660 mol of pentanol used

b)

mass of glacial acetic acid

Dglacial acetic acid = 1.05 g /mL

Mass = Density * Vlme = 1.05 *7 =7.35 g of glacial acetic acid

C)

moles of glacial acetic acid = mass/MW = 7.35/60.05 = 0.1224

d)

ratio is 1:1 so

pentanol has the LEAST amount of moles, 0.04660 mol vs. 0.1224 mol of glacial acetica acid

limiting = pentanol

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