Question

Please help me!!!!!!!!!!!!!

Answer 1

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:

$Al(C_2H_3O_2)_3 + 3NaOH --- > 3C_2H_3NaO_2 + Al(OH)_3$

The mole ratio of the 2 reactants is 1:3.

Mole of 100 mL. 1.25 mol/L $Al(C_2H_3O_2)_3$ = 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  $Al(C_2H_3O_2)_3$ and $Al(OH)_3$ = 1:1

Equivalent mole of $Al(OH)_3$ = 0.125 mol

Mass of 0.125 mole $Al(OH)_3$ = 0.125 x 78 = 9.75 grams.

More on stoichiometric calculations can be found here: brainly.com/question/27287858

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