Answer:
3.16g of the ester are produced
Explanation:
First, we need to determine the limiting reactant of the reaction converting each reactant to moles:
<em>Moles CH3COOH (Molar mass: 60g/mol):</em>
3.58g * (1mol / 60g) = 0.05967moles
<em>Moles C5H11OH (Molar mass: 88.15g/mol):</em>
4.75g * (1mol / 88.15g) = 0.05389 moles
As the reaction is 1:1, limiting reactant is C5H11OH.
Theoretical moles of the ester CH3COOC5H11 is 0.05389 moles.
As only the 45.00% is produced:
0.05389 moles * 45.00% =
0.02425 moles of CH3COOC5H11 are produced.
In mass (Molar mass CH3COOC5H11: 130.1849g/mol):
0.02425 moles * (130.1849g / mol) =
<h3>3.16g of the ester are produced</h3>