The complete reaction is as,
4-Aminophenol + Acetic Anhydride → <span>Acetaminophen + Acetic Acid
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>
Answer:
The number of moles of NaCl contained in 0.5000 L of a 1.5 M (Molarity) solution is 0.75 moles.
Explanation:
Molar concentration is a measure of the concentration of a solute in a solution, be it some molecular, ionic, or atomic species.
Molarity M is the number of moles of solute that are dissolved in a given volume.
The Molarity of a solution is determined by the following expression:
Molarity is expressed in units 
.
In this case:
- Molarity= 1.5 M
- number of moles of solute= ?
- volume= 0.5 L
Reemplacing:
Solving:
number of moles of solute= 1.5 M* 0.5 L
number of moles of solute= 0.75
<u><em>The number of moles of NaCl contained in 0.5000 L of a 1.5 M (Molarity) solution is 0.75 moles.</em></u>
Answer:
Ca3(PO4)2 and Na
Explanation:
The reaction equation is;
3Ca + 2Na3PO3 -------> Ca3(PO3)2 + 6Na
We must remember that calcium is above sodium in the electrochemical series. This implies that calcium can displace sodium from its solution. Calcium has a reduction potential of -2.87 V while sodium has a reduction potential of -2.71 V. The ion with the more negative reduction potential can easily displace the other in aqueous solution.
This explains why the reaction above is possible.
