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3 years ago

Ethanol (C2H5OH) is produced from the fermentation of sucrose in the presence of enzymes.

Chemistry

1 answer:

Klio2033 [76]3 years ago

5 0

Explanation:

to calculate theoratical yield...first we must calculate how much Sucrose moles we have..in order to to do so...we must calculate molar mass of sucrose.

molar mass of sucrose is 342.3 g/mol.

now we can calculate how many sucrose moles we have by dividing the mass with molar mass of sucrose

735g/(342.3g/mol)=2.147mol

theoratically...according to stoichiometry..every 1 mole of sucrose yields 4 moles of ethanol...so...

2.147mole yield 2.147*4mol=8.588mol

now we must calculate weight of that much ethanol..molar mass of ethanol is 46.07 g/mol...

so we can multiple moles by molar mass to obtain the weight 8.588mol*46.07g/mol=395.649g

but we only obtained 310.5g...so percentage we have is

$\frac{310.5}{395.649} \times 100$

78.47%

if there is trouble with molar mases I used....use what you calculated...

If my explanation is good enough...please mark it as brainliest.thanks

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Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

3 0

3 years ago

In the article "A Sea of Garbage" what message is each cartoon trying to convey to readers? Do they succeed?

sertanlavr [38]

Answer:

yes in the article I see of garbage what's messages each carton trying to convey to your readers with a success obviously they do suck said they succeed because like that's really great day socks said is your garbage is like a pool of garbage work garbage are gathered together they are together a you get one talking about you know when cabbage and together they make efficient use of them okay I mean garbage garbage garbage garbage so I don't know I don't know I don't know I don't know I don't know but it's really is regret it regret it was someone couldn't wish for our young people pick them up and see if garbage and it's really not making sense cuz when you see a sea of garbage that means there's water in there and people pull up garbage in be like water pollution or something but I'm actually not really like sure but I have a slight idea of like what we talking about here right now and it's great

8 0

3 years ago

What is the expected mass (in kg) of a 10 over 5 B isotope? 1 proton = 1.6726 × 10-27 kg 1 neutron = 1.6749 × 10-27 kg A.1.67 ×

Ira Lisetskai [31]

$^{10}_5 B$
An atom of this isotope contains 5 protons and 10-5=5 neutrons.

$5 \times 1.6726 \times 10^{-27} + 5 \times 1.6749 \times 10^{-27} = \\
8.363 \times 10^{-27} + 8.3745 \times 10^{-27}= \\
16.7375 \times 10^{-27}= \\
1.67375 \times 10^{-26} \approx \\
1.67 \times 10^{-26}$

The answer is A. 1.67 × 10⁻²⁶ kg.

8 0

4 years ago

Explain in a short sentence how you can tell a reaction is a decomposition reaction.

mr Goodwill [35]

Answer: A decomposition reaction occurs when one reactant breaks down into two or more products.

Mark me as brainilist pls

3 0

3 years ago

Calculate the amount of heat water absorbs from a piece of hot metal using the following data: 75.0 g of cold water is placed in

AysviL [449]

Answer:

Amount of heat absorbed by water is 2604.54 J.

Explanation:

Amount of heat absorbed by water = $m_{water}\times C_{water}\times \Delta T_{water}$

where m represents mass, C represents specific heat and $\Delta T$ represents change in temperature.

Here $m_{water}=75.0$ g , $C_{water}=4.184J/(g.^{0}\textrm{C})$ and $\Delta T$ = (final temperature - initial temperature) = (29.5-21.2) $^{0}\textrm{C}$ = 8.3 $^{0}\textrm{C}$

So, amount of heat heat absorbed by water

= $(75.0g)\times (4.184\frac{J}{g.^{0}\textrm{C}})\times (8.3^{0}\textrm{C})$

= 2604.54 J

8 0

4 years ago