1)The limiting reactant will be aluminum acetate
2) The mass of aluminum hydroxide formed will be 9.75 grams.
Stoichiometric problem
The equation of the reaction is as below:
[tex]Al(C_2H_3O_2)_3 + 3NaOH --- > 3C_2H_3NaO_2 + Al(OH)_3[/tex]
The mole ratio of the 2 reactants is 1:3.
Mole of 100 mL. 1.25 mol/L [tex]Al(C_2H_3O_2)_3[/tex] = 0.1 x 1.25 = 0.125 mol
Mole of 300 mL, 2.30 mol/L NaOH = 0.3 x 2.3 = 0.69 mol
Thus, aluminum acetate is limited.
Mole ratio of [tex]Al(C_2H_3O_2)_3[/tex] and [tex]Al(OH)_3[/tex] = 1:1
Equivalent mole of [tex]Al(OH)_3[/tex] = 0.125 mol
Mass of 0.125 mole [tex]Al(OH)_3[/tex] = 0.125 x 78 = 9.75 grams.
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