Question

a) Calculate the moles of t-penthl alcohol where 1.00 mL of t-pentyl alcohol (0.805 g/mol) is used.

b) Calculate the moles of 12 M HCl where 2.50 mL of 12 M HCL is used.

c) What is the limiting agent for the synthsis of t-pentyl chloride (by the reaxtion of t-pentyl alcohol woth concentrated hydrochloric acid)

d) Calculate the theoretical yield of t-pentyl chloride.

Thanks!

e) Calculate the percent yield of t-pentyl chloride.

Density: 0.805 g/cm−3

Mass of final product 50g

Answer 1

I calculated percent yield for 0.5 g instead of 50g since with 0.8 g pentyl alcohol 50g yield is may be correct. It may be 0.5g

- CHgctly of The equation of the reaction is shown below. CH3 CHg CH3 CHg CH OH + Hd . CHgch to Pentyl alcohol det t-Pentyl chloride molarmoss - 88 160 g/mol. 106.599/mo - volume of t-Pentyl alcohod used imL Density of t-pentyl alcohol = 0.8058/mL mass of t-pentyl alcohol used Density xodume = 0.8058 = 0.8058 ALIME Numbord mos 4 Pentyl alcohol mass : 0.8058-0-00913 may - nolar mass 88.150g/mol volume of Hd uscal = 2.50mL @ 0.0025L molarity of Hd = 12m. 14 moles Number of mole=nolarity xudume L volume in Litres rumber of moles of Hd used 12 mol/x0.00254=0.03 moles @ Pentyl alcohol is the limiting agent. Since its number of moles is 0.00913 modes is less than Hl Co.og moles). So the I Pentyl alcohol is limiting agent. molo mass product. Theoretical Yield mass of Reactant molea mass of Reactant = 0.8059 x 106.59 g/mol = 0.9749 Pentyl chloride - 0.9749 @ Experimental Xield = 50g . but this is not possible may be it is 500 ggch I am calculating Pencent xield for this field. Percent Xield of experimental xidd - 0.59 x 100 = 51.34% Pentyl chloride Theoretical xield to 0.9749 88.1584ml Theoretical Xieldq=0.9149