First of all convert the ml of Acetic anhydrite to grams, As, Density = mass / volume Solving for mass, mass = Density </span>× Volume <span>Putting values, mass = 1.08 g/ml </span>× 5ml <span> mass = 5.4 g of acetic anhydride
First Find amount of acetic anhydride required to react completely with 2 g of p-Aminophenol, As, 109.1 g of p-aminophenol required = 102.1 g of acetic anhydride so, 2 g of p-aminophenol will require = X g of Acetic Anhydride
Solving for X, X = (2 g </span>× 102.1 g) ÷ 109.1 g
X = 1.87 g of acetic anhydride is required to be reacted.
But, we are provided with 5.4 g of Acetic Anhydride, means p-aminophenol is the limiting reactant and it controls the formation of product. Now Let's calculate for product, As, 109.1 g of p-aminophenol produced = 180.2 g of <span>Acetaminophen So 2.00 g of p-aminophenol will produce = X g of Acetaminophen
Solving for X, X = (2.00 g </span>× 180.2 g) ÷ 109.1 g
X = 3.30 g of Acetaminophen
Result: <span>If 2.00g of p-aminophenol reacts with 5.00 ml of acetic anhydride 3.30 g of acetaminophen is made.</span>
Ionization energy (IE) is the amount of energy required to remove an electron.
If you observe the IEs sequentially, there is a large gap between the 2nd and 3rd. This suggests it is difficult to remove more than 2 two electrons. Elements that lose two electrons to become more stable are found in the Group 2A (2 representing the number of electrons in the outermost valence shell).
