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valentina_108 [34]
2 years ago
15

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

Chemistry
1 answer:
fenix001 [56]2 years ago
4 0

1)The limiting reactant will be aluminum acetate

2) The mass of aluminum hydroxide formed will be 9.75 grams.

<h3>Stoichiometric problem</h3>

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

#SPJ1

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