Question

4. NaOH react completely with Al(NO3)3.

a. Write the balance chemical equation

b. What volume of 0.300 M NaOH is required to react completely with 2.50 g Al(NO3)3

Answer 1

4.a.

Reaction between sodium hydroxide and aluminium nitrate can be given as follows

$3NaOH + Al(NO_3)_3\longrightarrow Al(OH)_3 +3 NaNO_3$

(b) 1 mole of Aluminium nitrate react with 3 mole NaOH ,

Mass of Aluminium nitrate is 2.50 g

Mole of aluminium nitrate = $\frac{2.5}{213}=0.0117$

mole of NaOH require to react completely with 0.0117 mole of aluminium nitrate = 0.035

$Molarity= \frac{mole}{volume}\times 1000= \frac{0.035}{V}\times 1000$

$0.3 = \frac{35}{V}$

V= 116.67 mL