72.1 g of steam is reacted with excess ethene (C2H4) according to the following reaction. Determine the maximum mass (in grams) of ethanol that could be produced.
C2H4 (g) + H2O (g) → C2H5OH (l)
mols H2O = g/molar mass = 72.1/18.01 = 4.0
Convert mols H2O to mols ethanol this way.
4.0 mols H2O x (1 mol ethanol/1 mol H2O) = 4 x /1/1 = 4.0 mols ethanol produced.
Then g ethanol = mols ethanol x molar mass ethanol.
Post your work if you get stuck.
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Why did the steam go on a date with ethene? Because they wanted to make some "high spirits" together! But let's see how much fun they had...
To determine the maximum mass of ethanol that could be produced, we need to find the limiting reactant - the reactant that will run out first. In this case, we know that we have 72.1 g of steam, but the amount of ethene is not given. So, let's calculate it together!
The molar mass of C2H4 is 28.05 g/mol, and the molar mass of H2O is 18.015 g/mol. To find the amount of ethene, we'll need to convert 72.1 g of steam to moles:
72.1 g H2O × (1 mol H2O/18.015 g H2O) = 4.00 mol H2O
The balanced equation tells us that the ratio between C2H4 and H2O is 1:1. Therefore, the amount of ethene (C2H4) will also be 4.00 mol.
Now, we need to convert the amount of C2H4 in moles to mass:
4.00 mol C2H4 × (28.05 g C2H4/ 1 mol C2H4) = 112 g C2H4
So, if we react 72.1 g of steam, the maximum mass of ethanol that could be produced is 112 g. That's quite a party!
To determine the maximum mass of ethanol that could be produced, we need to use stoichiometry. Here's how you can calculate it:
Step 1: Write a balanced equation for the reaction, and determine the stoichiometric ratios.
C2H4 (g) + H2O (g) → C2H5OH (l)
From the balanced equation, we can see that 1 mole of C2H4 reacts with 1 mole of H2O to produce 1 mole of C2H5OH.
Step 2: Convert grams of steam (H2O) to moles.
To do this, we need to use the molar mass of water. The molar mass of H2O is approximately 18.015 g/mol.
Mass of H2O (g) = 72.1 g
Moles of H2O = Mass of H2O (g) / Molar mass of H2O (g/mol)
Moles of H2O = 72.1 g / 18.015 g/mol
Moles of H2O = 4.00 mol
Step 3: Determine the moles of C2H5OH produced.
Since the stoichiometric ratio between H2O and C2H5OH is 1:1, the moles of H2O also represent the moles of C2H5OH produced.
Moles of C2H5OH = Moles of H2O
Moles of C2H5OH = 4.00 mol
Step 4: Convert moles of C2H5OH to grams.
To do this, we need to use the molar mass of C2H5OH. The molar mass of C2H5OH is approximately 46.07 g/mol.
Mass of C2H5OH (g) = Moles of C2H5OH × Molar mass of C2H5OH (g/mol)
Mass of C2H5OH (g) = 4.00 mol × 46.07 g/mol
Mass of C2H5OH (g) = 184.28 g
Therefore, the maximum mass of ethanol that could be produced is 184.28 grams.