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!

Answer 1

Here's what I got.

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"#

Use the molar mass of the compound to convert this to moles.

#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, use the element's molar mass to convert the mass of magnesium to moles.

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

Likewise with 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 are aware that you have to have

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

At this stage, you have to choose one of the three reactants and determine whether you have enough moles of the other two to guarantee that that reactant reacts in full.

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 an adequate supply of magnesium in moles?

#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 methyl benzoate in sufficient moles on hand?

#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 indicates that before all of the moles of magnesium and bromobenzene have a chance to react, all of the methyl benzoate will be consumed.

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"#

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Answer 2

Kai Bowman

The limiting reagent in the formation of triphenylmethanol depends on the specific reaction conditions and the amounts of reactants used. It is determined by comparing the stoichiometric ratios of the reactants involved in the reaction.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.