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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1. Fructose
2. Proteins
3. Glucose
4. Ribose
5. Galactose.
This is also called sugar, or carbohydrates.
Answer: put 2 in front of NaOH and 2 also in front of H2O
Explanation: 2NaOH + H2CO3 —> Na2CO3 + 2H2O
We can find it by 2n^2 formula
So
First shell is maximum of 2
Second shell is of 8
Third shell is of 18
Fourth shell is of 32
Gravity depends on inertia