C2H4(g) + H2O(l)
→ C2H5O(l)
what is the rendition percentage if 4.50g of C2H4 produce 4.7g of ethyl alcohol?
I have no idea what a "rendition percent" is. I'm skeptical that the product is C2H5O. I think it may be C2H6O
Finish (if necessary) and balance the equation.
Convert 450 g of C2H4 to mols. mols = grams/molar mass.
Use the coefficients in the balanced equation to convert mols C2H4 to the product.
Convert mols product to grams. g = mols x molar mass. This is the theoretical yield (TY). The actual yield (AY) is 4.7g.
% yield = (AY/TY)*100 = ?
To calculate the percentage yield for this reaction, you need to compare the actual yield (4.7g) with the theoretical yield. The theoretical yield is the maximum amount of product that can be obtained from the given amount of reactant.
First, let's calculate the molar mass of C2H4 (ethylene):
C2H4: 2 * (12.01 g/mol) + 4 * (1.01 g/mol) = 28.05 g/mol
Next, calculate the moles of C2H4 by dividing the given mass (4.50g) by the molar mass:
moles of C2H4 = mass / molar mass = 4.50g / 28.05 g/mol ≈ 0.1603 mol
The balanced equation shows that 1 mole of C2H4 reacts to produce 1 mole of C2H5O (ethyl alcohol). Therefore, the moles of C2H5O produced is also approximately 0.1603 mol.
Now, calculate the theoretical yield of ethyl alcohol (C2H5O) using its molar mass:
C2H5O: 2 * (12.01 g/mol) + 6 * (1.01 g/mol) + 16.00 g/mol = 46.08 g/mol
theoretical yield = moles of C2H5O * molar mass = 0.1603 mol * 46.08 g/mol ≈ 7.391 g
Finally, calculate the percentage yield:
percentage yield = (actual yield / theoretical yield) * 100%
= (4.7g / 7.391g) * 100%
≈ 63.5%
Therefore, the percentage yield of the reaction is approximately 63.5%.