Given each pair, complete the sentences to determine which member of each has the stronger intermolecular dispersion forces.
1 answer:
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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:
The mole ratio of the 2 reactants is 1:3.
Mole of 100 mL. 1.25 mol/L = 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 and
= 1:1
Equivalent mole of = 0.125 mol
Mass of 0.125 mole = 0.125 x 78 = 9.75 grams.
More on stoichiometric calculations can be found here: brainly.com/question/27287858
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