Question

Question #971ca

Answer 1

The limiting reagent will be nitric acid.

Explanation:

Your question is incomplete. In order to be able to calculate the percent yield of the reaction you need to know the actual amount of nitroglycerin produced.

As it is written, the question only allows you to determine the limiting reagent and how much of the excess reagent is left after the reaction takes place.

Start from the balanced chemical equation for the reaction between glycerol and nitric acid

`"C"_3"H"_5("OH")_3 + color(red)(3)"HNO"_3 -> "C"_3"H"_5("NO")_3 + 3"H"_2"O"`

The `1:color(red)(3)` mole ratio that exists between glycerol and nitric acid tells you that, regardless of how many moles of the former you have, the reaction needs three times more moles of the latter.

Use glycerol and nitric acid's molar masses to determine how many moles of each you have

`20color(red)(cancel(color(black)("g"))) * "1 mole glycerol"/(92.09color(red)(cancel(color(black)("g")))) = "0.217 moles glycerol"`

and

`10color(red)(cancel(color(black)("g"))) * "1 mole nitric acid"/(63.01color(red)(cancel(color(black)("g")))) = "0.159 moles nitric acid"`

It's clear that nitric acid will be the limiting reagent, since you actually have less of it present. This means that not all the gycerol will react.

More specifically, the reaction will only consume

`0.159color(red)(cancel(color(black)("moles HNO"""_3))) * ("1 mole C"_3"H"_5("OH")_3)/(color(red)(3)color(red)(cancel(color(black)("moles HNO"""_3)))) = "0.053 moles C"""_3"H"_5("OH")_3`

This means that you have an excess of

`n_"excess" = 0.217 - 0.053 = "0.164 moles C"""_3"H"_5("OH")_3`

This corresponds to a mass of

`0.164color(red)(cancel(color(black)("moles"))) * "92.09 g"/(1color(red)(cancel(color(black)("mole")))) = "15.1 g of glycerol"`

The `1:1` mole ratio that exists between glycerol and nitroglycerin will help you determine the theoretical yield of the reaction, which is what you'll get for a percent yield of 100%.

`0.053color(red)(cancel(color(black)("moles C"""_3"H"_5("OH")_3))) * ("1 mole C"_3"H"_5("NO")_3)/(1color(red)(cancel(color(black)("mole C"""_3"H"_5("OH")_3)))) = "0.053 moles C"""_3"H"_5("NO")_3`

Use the molar mass of nitroglycerin to calculate the mass produced

`0.053color(red)(cancel(color(black)("moles"))) * "227.09 g"/(1color(red)(cancel(color(black)("mole")))) = "12.0 g"`

Assuming that your reaction actually produced `x` grams of nitroglycerin, the percent yield would be equal to

`"% yield" = "actual yield"/"theoretical yield" * 100`

`"% yield" = (xcolor(red)(cancel(color(black)("g"))))/(12.0color(red)(cancel(color(black)("g")))) * 100 = 8.33 * x`

So, for example, if your reaction produced 8.5 g of nitroglycerin, the percent yield of the reaction was

`"% yield" = 8.33 * 8.5 = "70.8%"`