If 250g of sugar is completely fermented to ethanol, what is the theoretical yield of ethyl alcohol in: ?

i. grams or milliliters (density of ethanol is 0.79g/mL)

ii. what is the final %v/v of alcohol if that 250 sugar was used to make 1L of wine? (assume no volume changed during fermentation)

please, show me the working out.

Thanks

Answer 1

i. The theoretical yield of ethanol is 135 g or 170 mL; ii.the final concentration is 17 % v/v.

Let's compile all the data in one location since we know we'll need a balanced equation with masses and moles.

I assume that by "sugar" you mean sucrose, #"C"_12"H"_22"O"_11#.
#M_text(r):color(white)(mmm)342.30color(white)(mmmmmmmmll)46.07# #color(white)(mmmm)"C"_12"H"_22"O"_11 + "H"_2"O" → 4"CH"_3"CH"_2"OH" + 4"CO"_2# #"Mass/g:"color(white)(mll)250#

Determine the moles of sugar in step 1.

#"Moles of sugar" = 250 color(red)(cancel(color(black)("g sugar"))) × "1 mol sugar"/(342.30 color(red)(cancel(color(black)("g sugar")))) = "0.7304 mol sugar"#

Step 2: Determine the ethanol moles.

#"Moles of ethanol" = 0.7304 color(red)(cancel(color(black)("mol sugar"))) × "4 mol ethanol"/(1 color(red)(cancel(color(black)("mol sugar")))) = "2.921 mol ethanol"#

Step 3: Determine the ethanol's mass.

#"Mass of ethanol" = 2.921 color(red)(cancel(color(black)("mol ethanol"))) × "46.07 g ethanol"/(1 color(red)(cancel(color(black)("mol ethanol")))) = "135 g ethanol"#

Step 4: Determine the ethanol volume.

#"Volume of ethanol" = 135 color(red)(cancel(color(black)("g ethanol"))) × "1 mL ethanol"/(0.79 color(red)(cancel(color(black)("g ethanol")))) = "170 mL ethanol"#

Step 5: Determine the ethanol volume percentage.

#"Volume %" = "volume of ethanol"/"total volume" × 100 % = (170 color(red)(cancel(color(black)("mL"))))/(1000 color(red)(cancel(color(black)("mL")))) × 100 % = "17 % v/v"#
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Answer 2

Here's what I got.

Start by writing the chemical equation that describes this reaction

#"C"_ 6"H"_ 12"O"_ (6(aq)) -> 2"C"_ 2"H"_ 5"OH"_ ((aq)) + 2"CO"_ (2(g)) uarr#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)# SIDE NOTE As you can see, I am assuming that by #"250 g"# of sugar you mean #"250 g"# of glucose, not sucrose, #"C"_12"H"_22"O"_11#.

If you were supposed to work with sucrose instead of glucose, add this step

#"C"_ 12"H"_ 22"O"_ (11(aq)) + "H"_ 2"O"_ ((l)) -> 2"C"_ 6"H"_ 12 "O"_ (6(aq))#
and redo the calculations with the mass of glucose that you'll get from this reaction. #color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#
Now, notice that you have a #1:2# mole ratio between glucose and ethanol, so convert the mass of glucose to moles by using the compound's molar mass
#250 color(red)(cancel(color(black)("g"))) * "1 mole glucose"/(180.156color(red)(cancel(color(black)("g")))) = "1.388 moles glucose"#

This means that the reaction will produce

#1.388 color(red)(cancel(color(black)("moles glucose"))) * "2 moles ethanol"/(1color(red)(cancel(color(black)("mole glucose")))) = "2.776 moles ethanol"#

To convert this to grams, use the molar mass of ethanol

#2.776 color(red)(cancel(color(black)("moles ethanol"))) * "46.07 g"/(1color(red)(cancel(color(black)("mole ethanol")))) = "127.9 g"#

Rounded to two sig figs, the answer will be

#color(darkgreen)(ul(color(black)("theoretical yield = 130 g")))#

To convert this to milliliters, use the density of ethanol

#127.9 color(red)(cancel(color(black)("g"))) * "1 mL"/(0.79color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("160 mL")))#
To find the volume by volume percent concentration of the wine, #"% v/v"#, you should assume that you're working with the volume of ethanol produced by the reaction in a total volume of #"1 L" = 10^3# #"mL"# of wine.
Now, the solution's volume by volume percent concentration tells you the volume of solute present for every #"100 mL"# of solution.
In your case, #"100 mL"# of your solution will contain
#100 color(red)(cancel(color(black)("mL solution"))) * "160 mL ethanol"/(10^3color(red)(cancel(color(black)("mL solution")))) = "16 mL ethanol"#

Therefore, the concentration by volume will be

#color(darkgreen)(ul(color(black)("% v/v = 16% ethanol")))#

I'll leave the answer rounded to two sig figs, but don't forget that you only have one significant figure for the volume of wine.

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Answer 3

To determine the theoretical yield of ethanol produced from the fermentation of 250g of sugar, we need to use the stoichiometry of the reaction. The balanced chemical equation for the fermentation of sugar (glucose) to ethanol and carbon dioxide is:

C6H12O6 → 2 C2H5OH + 2 CO2

From the balanced equation, we can see that 1 mole of glucose produces 2 moles of ethanol. The molar mass of glucose (C6H12O6) is approximately 180 g/mol.

First, we need to convert the mass of sugar (250g) to moles: 250g / 180 g/mol = 1.39 moles of sugar

Since the molar ratio of sugar to ethanol is 1:2, we multiply the number of moles of sugar by 2 to find the number of moles of ethanol produced: 1.39 moles of sugar × 2 = 2.78 moles of ethanol

Now, we convert the moles of ethanol to grams using the molar mass of ethanol (C2H5OH), which is approximately 46 g/mol: 2.78 moles × 46 g/mol = 127.88g

Therefore, the theoretical yield of ethanol produced from the fermentation of 250g of sugar is approximately 127.88 grams.

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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.

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.

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.

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.

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