Answer:
81.97 g of NaAl(OH)₄
Solution:
The reaction for the preparation of Sodium Aluminate from Aluminium Metal and NaOH is as follow,
2 NaOH + 2 Al + 6 H₂O → 2 NaAl(OH)₄ + 3 H₂
According to this equation,
2 Moles of NaOH produces = 163.94 g (2 mole) of NaAl(OH)₄
So,
1 Mole of NaOH will produce = X g of NaAl(OH)₄
Solving for X,
X = (1 mol × 163.94 g) ÷ 2 mol
X = 81.97 g of NaAl(OH)₄
Answer:
0.0970 M
Explanation:
Remember this equation:
mol/M x V
Convert it so that you can get M.
M=mol/V
Convert the 2.14 grams of H2SO4 into mols
=0.0218
Convert mL to L
225/1000
=0.225
Plug it in.
0.0218/0.225
=0.0970 M
According to Bronsted-Lowry theory of acids and bases, an acid is proton donor and a base is proton acceptor. When an acid loses the proton, it forms the conjugate base of the acid. Similarly when a base accepts a proton, it forms the conjugate acid if that base. A conjugate acid base pair differs by a single proton.
Given the weak acid
. It can lose a proton to form the conjugate base.

Therefore, the conjugate base of
is
.
To balance the redox equation in the ion-electron method (also called the half-reaction method), the redox equation is separated into two half-equations - one for oxidation and one for reduction. Each of these half-reactions are balanced separately by the number of electrons which accepted by reduction or donated by oxidation and then combined to give the balanced redox equation.