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
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
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