Answer:
7.5 g of AlCl3
Explanation:
The given equation is;
NaOH + AlCl3 --> Al(OH)3 + NaCI.
By inspection, it is not balanced because OH and Clare not equal on both sides of the equation.
Thus, let's make them equal by balancing the equation.
Cl has 3 on the left, so we will make it to have 3 on the right. Same thing with OH on the right and we will make it to have 3 on the left. Thus:
3NaOH + AlCl3 --> Al(OH)3 + 3NaCI
We can see that;
NaOH has 3 moles
While AlCl3 has 1 mole
Thus, to find how many grams of AlCl3 will be required to completely react with 2.25g of NaOH ;
2.25g of NaOH × (3 moles NaOH/39.997 g/mol of NaOH) × (1 mole of AlCl3/3 moles of NaOH) × (133.34 g/mol of AlCl3/1 mol AlCl3) = 7.5 g of AlCl3
The addition of hydrofluoric acid and Naf( answer c) to water produces a buffer solution.
Naf is used because buffer consist of a weak acid and the salt of that base.Buffer also consist a weak base and salt of that base .Hydrofluoric acid is a weak acid thus the salt would be added is Naf to produce buffer solution.
