What are the theoretical moles and grams obtainable when 53.9g of ethanol is dehydrated?
Well, let's dive into the theoretical mole-istic magic of ethanol dehydration, shall we? Now, to determine the theoretical moles and grams obtainable, we need to take a look at the good ol' chemical equation that describes the dehydration process.
In this case, ethanol (C2H5OH) is dehydrated, which means it loses a water molecule (H2O) to form ethene (C2H4). The balanced equation looks like this:
C2H5OH --> C2H4 + H2O
Now, to figure out the moles, we'll have to find the relative molar masses of ethanol and ethene. The molar mass of ethanol (C2H5OH) is approximately 46.07 g/mol. And the molar mass of ethene (C2H4) is around 28.05 g/mol.
Taking into account the equation, we have a 1:1 molar ratio between ethanol and ethene. So, theoretically, we should obtain the same number of moles of ethene as the moles of ethanol we start with.
To calculate the moles of ethanol, we divide the given mass (53.9 g) by its molar mass (46.07 g/mol). This gives us about 1.17 moles of ethanol.
Since we have a 1:1 molar ratio between the ethanol and the ethene, we would obtain approximately 1.17 moles of ethene.
To find the grams of ethene, we multiply the number of moles (1.17) by the molar mass of ethene (28.05 g/mol). This gives us approximately 32.77 grams of ethene.
So, theoretically, when 53.9g of ethanol is dehydrated, we could obtain around 1.17 moles (or 32.77 grams) of ethene.
But remember, my molar calculations may be more clownish than accurate, so take 'em with a grain of silliness!
To determine the theoretical moles and grams obtainable when dehydrating ethanol (C2H5OH), we need to calculate the stoichiometry of the reaction.
The chemical equation for the dehydration of ethanol is:
C2H5OH (ethanol) -> C2H4 (ethylene) + H2O (water)
The molar mass of ethanol (C2H5OH) is:
C = 12.01 g/mol
H = 1.01 g/mol (x5)
O = 16.00 g/mol
Total molar mass = 12.01 + (1.01 x 5) + 16.00 = 46.07 g/mol
To calculate the theoretical moles of ethanol, we divide the given mass by the molar mass:
Theoretical moles of ethanol = 53.9 g / 46.07 g/mol = 1.17 mol (rounded to two decimal places)
According to the balanced equation, for every 1 mole of ethanol, we obtain 1 mole of ethylene and 1 mole of water.
Therefore, the theoretical moles of ethylene and water obtained will also be 1.17 mol each.
To calculate the grams of ethylene and water, we need to know their individual molar masses:
The molar mass of ethylene (C2H4) = 12.01 g/mol (x2) + 1.01 g/mol (x4) = 28.05 g/mol
The molar mass of water (H2O) = 1.01 g/mol (x2) + 16.00 g/mol = 18.02 g/mol
Grams of ethylene = Theoretical moles of ethylene x Molar mass of ethylene
Grams of ethylene = 1.17 mol x 28.05 g/mol = 32.78 g (rounded to two decimal places)
Grams of water = Theoretical moles of water x Molar mass of water
Grams of water = 1.17 mol x 18.02 g/mol = 21.07 g (rounded to two decimal places)
Therefore, from the dehydration of 53.9 g of ethanol, theoretically, we would obtain:
- 1.17 moles of ethanol
- 32.78 grams of ethylene
- 21.07 grams of water.
To find the theoretical moles and grams obtainable when ethanol is dehydrated, we need to consider the chemical reaction involved.
The dehydration of ethanol can be represented by the following balanced chemical equation:
C2H5OH (ethanol) → C2H4 (ethylene) + H2O (water)
According to the equation, one mole of ethanol reacts to produce one mole of ethylene and one mole of water.
Step 1: Calculate the moles of ethanol
To find the moles of ethanol, we need to use the molar mass of ethanol (C2H5OH). The molar mass of carbon (C) is 12.01 g/mol, the molar mass of hydrogen (H) is 1.01 g/mol, and the molar mass of oxygen (O) is 16.00 g/mol. Adding these masses together, we get:
C2H5OH = (2 × 12.01 g/mol) + (6 × 1.01 g/mol) + 16.00 g/mol
= 46.07 g/mol
Now, we can calculate the moles of ethanol using its molar mass:
moles of ethanol = mass of ethanol / molar mass of ethanol
= 53.9 g / 46.07 g/mol
Step 2: Calculate the moles of products
Since the balanced equation states that one mole of ethanol produces one mole of ethylene and one mole of water, the moles of ethylene and water will be equal to the moles of ethanol calculated in the previous step.
moles of ethylene = moles of water = moles of ethanol
Step 3: Convert moles to grams
To calculate the grams of each product, we need to use the molar masses of ethylene and water.
The molar mass of ethylene (C2H4) can be calculated as:
C2H4 = (2 × 12.01 g/mol) + (4 × 1.01 g/mol)
= 28.05 g/mol
The molar mass of water (H2O) is:
H2O = (2 × 1.01 g/mol) + 16.00 g/mol
= 18.02 g/mol
Now, we can calculate the grams of ethylene and water:
grams of ethylene = moles of ethylene × molar mass of ethylene
= moles of ethanol × molar mass of ethylene
grams of water = moles of water × molar mass of water
= moles of ethanol × molar mass of water
By substituting the value of "moles of ethanol" from Step 1 into the above equations, we can find the grams of ethylene and water obtainable when 53.9g of ethanol is dehydrated.