Which of these is the limiting reagent in the formation of triphenylmethanol?

Reacting to form triphenylmethanol (MW=260.33):

2.5 mL of bromobenzene (MW= 157.01, Density=1.491)
0.505g of solid magnesium (MW=24)
1.2g methylbenzoate (MW=136, Density=1.09)

Stoichiometric ratios:
Bromobenzene:Mg:Methylbenzoate:Triphenylmethanol = 2:1:1:1

My TA said it will be bromobenzene, but I can't justify it in my calculations. I must be doing something wrong, but I need help finding what it is, and my TA is on holiday.

Thanks!

1 Answer
Mar 17, 2017

Answer:

Here's what I got.

Explanation:

As you know, the idea here is that you need to compare the number of moles of each reactant to the #2:1:1# mole ratio you have for this reaction.

Let's start with bromobenzene, #"C"_6"H"_5"Br"#. You know that your sample has a volume of #"2.5 mL"# and a density of #"1.491 g mL"^(-1)#, which means that it has a mass of

#2.5 color(red)(cancel(color(black)("mL"))) * "1.491 g"/(1color(red)(cancel(color(black)("mL")))) = "3.7275 g"#

To convert this to moles, use the compound's molar mass

#3.7275 color(red)(cancel(color(black)("g"))) * ("1 mole C"_6"H"_5"Br")/(157.01color(red)(cancel(color(black)("g")))) = "0.02374 moles C"_6"H"_5"Br"#

Next, convert the mass of magnesium to moles by using the element's molar mass

#0.505 color(red)(cancel(color(black)("g"))) * "1 mole Mg"/(24color(red)(cancel(color(black)("g")))) = "0.02104 g Mg"#

Do the same for methyl benzoate

#1.2 color(red)(cancel(color(black)("g"))) * ("1 mole C"_6"H"_5"CO"_2"CH"_3)/(136color(red)(cancel(color(black)("g")))) = "0.008824 moles C"_6"H"_5"CO"_2"CH"_3#

You know that you must have

#"C"_6"H"_5"Br " : " Mg " : " C"_6"H"_5"CO"_2"CH"_3 = 2: 1 : 1#

At this point, you must pick one of the three reactants and check to see if you have enough moles of the other two to ensure that all the moles of said reactant take part in the reaction.

Let's pick bromobenzene. You can say that #0.02374# moles of bromobenzene would require

#0.02374 color(red)(cancel(color(black)("moles C"_6"H"_5"Br"))) * "1 mole Mg"/(2color(red)(cancel(color(black)("moles C"_6"H"_5"Br")))) = "0.01187 moles Mg"#

Do you have enough moles of magnesium available?

#overbrace("0.02104 moles Mg")^(color(blue)("what you have")) " ">" " overbrace("0.01187 moles Mg")^(color(purple)("what you need"))#

so yes, you do #-># magnesium is not the limiting reagent.

#0.02374 color(red)(cancel(color(black)("moles C"_6"H"_5"Br"))) * ("1 mole C"_6"H"_5"CO"_2"CH"_3)/(2color(red)(cancel(color(black)("moles C"_6"H"_5"Br"))))#

# = "0.01187 moles C"_6"H"_5"CO"_2"CH"_3#

Do you have enough moles of methyl benzoate available?

#overbrace("0.008824 moles C"_6"H"_5"CO"_2"CH"_3)^(color(blue)("what you have"))" " <" " overbrace("0.01187 moles C"_6"H"_5"CO"_2"CH"_3)^(color(purple)("what you need"))#

so no, you don't #-># methyl benzoate will be the limiting reagent.

This means that methyl benzoate will be completely consumed before all the moles of bromobenzene and magnesium will get the chance to react.

More specifically, the reaction will consume #0.008824# moles of methyl benzoate,

#0.008824 color(red)(cancel(color(black)("moles C"_6"H"_5"CO"_2"CH"_3))) * "1 mole Mg"/(1color(red)(cancel(color(black)("mole C"_6"H"_5"CO"_2"CH"_3))))#

# = " 0.008824 moles Mg"#

and

#0.008824 color(red)(cancel(color(black)("moles C"_6"H"_5"CO"_2"CH"_3))) * ("2 moles C"_6"H"_5"Br")/(1color(red)(cancel(color(black)("mole C"_6"H"_5"CO"_2"CH"_3))))#

# = "0.01765 moles C"_6"H"_5"Br"#