Question #19e99

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Question #19e99

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Answer 1 · 2014-01-15T01:37:11

It is almost certainly the salicylic acid.

Explanation:

You usually use the cheapest reagents as the excess reactants.

Sulfuric acid and acetic anhydride are dirt-cheap.

However, you have to do the calculation using the quantities you used in your experiment.

EXAMPLE

You prepared aspirin by reacting 2.00 g salicylic acid, 4.00 mL acetic anhydride (density = 1.08 g/mL), and a few drops of concentrated sulfuric acid. What was the limiting reactant?

Solution

You need to write the balanced equation for the reaction.

${({CH}_{3} CO)}_{2} O + {HOC}_{6} H_{4} COOH \to {CH}_{3} {COOC}_{6} H_{4} COOH + {CH}_{3} COOH$ $("CH"_3"CO")_2"O" + "HOC"_6"H"_4"COOH" → "CH"_3"COOC"_6"H"_4"COOH" + "CH"_3"COOH"$

$C_{4} H_{6} O_{3} + C_{7} H_{6} O_{3} \to C_{9} H_{8} O_{4} + C_{2} H_{4} O_{2}$ $"C"_4"H"_6"O"_3 + "C"_7"H"_6"O"_3 → "C"_9"H"_8"O"_4 + "C"_2"H"_4"O"_2$

For simplicity, let’s rewrite this as

${Ac}_{2} O + SA \to Asp + AcOH$ $"Ac"_2"O" + "SA" → "Asp" + "AcOH"$

The sulfuric acid is a catalyst (not used up), so we need to consider only the ${Ac}_{2} O$ $"Ac"_2"O"$ and $SA$ $"SA"$ as possible limiting reactants.

We know that we will need a balanced equation, molar masses, and moles of the compounds involved.

1. Gather all the information in one place with the molecular masses above the formulas and the masses and moles below them.

${Mass of Ac}_{2} O = 4.00 {mL Ac}_{2} O \times \frac{{1.08 g Ac}_{2} O}{1 {mL Ac}_{2} O} = {4.32 g Ac}_{2} O$ $"Mass of Ac"_2"O" = 4.00 color(red)(cancel(color(black)("mL Ac"_2"O"))) × ("1.08 g Ac"_2"O")/( 1 color(red)(cancel(color(black)("mL Ac"_2"O")))) = "4.32 g Ac"_2"O"$

$M_{r} : m m m l l 102.09 m m l 138.12 m m l 180.16$ $M_"r":color(white)(mmmll)102.09color(white)(mml)138.12color(white)(mml)180.16$
$m m m m m m {Ac}_{2} O m + m SA m \to m Asp + AcOH$ $color(white)(mmmmmm)"Ac"_2"O"color(white)(m) +color(white)(m) "SA"color(white)(m) →color(white)(m) "Asp" + "AcOH"$
$Mass/g: m m l 4.32 m m m l l 2.00$ $"Mass/g:"color(white)(mml)4.32color(white)(mmmll)2.00$
$Amt/mol: l l 0.042 32 m l 0.014 48$ $"Amt/mol:"color(white)(ll)"0.042 32"color(white)(ml)"0.014 48"$
$Divide by: m m l l 1 m m m m l 1$ $"Divide by:"color(white)(mmll)1color(white)(mmmml)1$
$Moles rxn: l l 0.042 32 m 0.014 48$ $"Moles rxn:"color(white)(ll)"0.042 32"color(white)(m)"0.014 48"$

2. Identify the limiting reactant

An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will give:

You divide the moles of each reactant by its corresponding coefficient in the balanced equation.

I did that for you in the table above.

$SA$ $"SA"$ is the limiting reactant because it gives fewer moles of reaction.

The ${Ac}_{2} O$ $"Ac"_2"O"$ was used in excess to make sure that the maximum amount of $SA$ $"SA"$ was converted to aspirin.

Here's a video that shows the "usual" way of finding a limiting reactant.

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Question #19e99

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