How many grams of c6h12o6 are needed to form 7.50g of c2h5oh
THINK IT"S
Molar mass C2H5OH = 46 g / mol
7.50 g / 46 = 0.163 moles
the ratio between C6H12O6 and C2H5OH is 1 : 2
1 : 2 = x : 0.163
x = 0.0815 moles of C6H12O6 needed
molar mass C6H12O6 = 180 g/mol
180 x 0.0815 = 14.7 g
Well, let's break it down! C6H12O6 is the chemical formula for glucose, and C2H5OH is the formula for ethanol. To figure out the grams of C6H12O6 needed, we need to do some stoichiometry.
So, without further ado, let's put on our chemist's clown shoes and get this formula circus started!
First off, we need to know the molar mass of C2H5OH and C6H12O6. The molar mass of C2H5OH is approximately 46.07 g/mol, and the molar mass of C6H12O6 is approximately 180.16 g/mol.
Now, we can set up a proportion to find the grams of C6H12O6 needed:
(180.16 g C6H12O6 / 1 mol C6H12O6) = (7.50 g C2H5OH / x)
Cross multiplying, we get:
180.16 g C6H12O6 * x = 7.50 g C2H5OH
Solving for x, we find:
x = 7.50 g C2H5OH / 180.16 g C6H12O6
Calculating, we have:
x ≈ 0.0416 mol C6H12O6
Finally, we can convert this to grams of C6H12O6:
0.0416 mol C6H12O6 * 180.16 g C6H12O6 / 1 mol C6H12O6 ≈ 7.50 g C6H12O6
So, approximately 7.50 grams of C6H12O6 are needed to form 7.50 grams of C2H5OH!
And there you have it, a little chemistry and a lot of clowning around!
To determine the number of grams of C6H12O6 needed to form 7.50g of C2H5OH, we need to set up a balanced chemical equation and use stoichiometry.
The balanced equation for the reaction between C6H12O6 (glucose) and C2H5OH (ethanol) is:
C6H12O6 -> 2 C2H5OH
The molar mass of C6H12O6 is:
(6 * 12.01 g/mol) + (12 * 1.01 g/mol) + (6 * 16.00 g/mol) = 180.18 g/mol
Using the molar mass of C6H12O6, we can calculate the number of moles of C6H12O6 needed to produce 7.50g of C2H5OH:
Number of moles of C6H12O6 = (mass of C2H5OH / molar mass of C2H5OH) * (1 mole C6H12O6 / 2 moles C2H5OH)
Number of moles of C6H12O6 = (7.50g / 46.07 g/mol) * (1 mole C6H12O6 / 2 moles C2H5OH)
= 0.1626 moles C6H12O6
Finally, we can calculate the mass of C6H12O6 needed using the molar mass of C6H12O6:
Mass of C6H12O6 = number of moles of C6H12O6 * molar mass of C6H12O6
= 0.1626 moles * 180.18 g/mol
= 29.32 grams
Therefore, approximately 29.32 grams of C6H12O6 are needed to form 7.50 grams of C2H5OH.
To calculate the number of grams of C6H12O6 needed to form 7.50g of C2H5OH (ethanol), we first need to understand the balanced chemical equation for the reaction.
The balanced equation for the fermentation of glucose (C6H12O6) to form ethanol (C2H5OH) is:
C6H12O6 → 2 C2H5OH + 2CO2
From the equation, you can see that 1 mole of glucose (C6H12O6) reacts to form 2 moles of ethanol (C2H5OH). Now we can proceed with the calculations:
1. Calculate the molar mass of C6H12O6 (glucose):
C: 6 x 12.01 g/mol = 72.06 g/mol
H: 12 x 1.008 g/mol = 12.096 g/mol
O: 6 x 16.00 g/mol = 96.00 g/mol
Total molar mass of C6H12O6: 72.06 + 12.096 + 96.00 = 180.156 g/mol
2. Calculate the moles of C2H5OH (ethanol) using the given mass:
Moles = Mass / Molar mass
Moles = 7.50 g / 46.07 g/mol (molar mass of C2H5OH)
Moles ≈ 0.163 moles (rounded to three decimal places)
3. Use the stoichiometry of the balanced equation to determine the moles of C6H12O6 needed:
According to the equation, 1 mole of C6H12O6 reacts to form 2 moles of C2H5OH.
Therefore, 0.163 moles of C2H5OH will need half as many moles of C6H12O6:
Moles of C6H12O6 needed = 0.163 moles / 2 = 0.0815 moles (rounded to four decimal places)
4. Finally, calculate the grams of C6H12O6 needed:
Mass = Moles × Molar mass
Mass = 0.0815 moles × 180.156 g/mol (molar mass of C6H12O6)
Mass ≈ 14.71 g (rounded to two decimal places)
Therefore, approximately 14.71 grams of C6H12O6 are needed to form 7.50 grams of C2H5OH.