Question

quien y en que cantidad sera el reactivo limite, si utilizamos 125 g de ácido, H (CH3COO) y 275 g de hidróxido , Al(OH)3

Answer 1

Answer:

The limiting reactant is acetic acid. All 125 g will react.  

Explanation:

1. Assemble the information

We will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.  

Mᵣ:                 60.05            78.00

                3CH₃COO-H + Al(OH)₃ ⟶ (CH₃COO)₃Al + 3H₂O

Mass/g:           125               275  

2. Calculate the moles of each reactant

$\text{Moles of CH _{3} COOH} = \text{125 g CH _{3} COOH} \times \dfrac{\text{1 mol CH _{3} COOH}}{\text{60.05 g CH _{3} COOH}} = \text{2.082 mol CH _{3} COOH }\\\\\text{Moles of Al(OH)}_{3} = \text{275 g Al(OH)}_{3} \times \dfrac{\text{1 mol Al(OH)}_{3}}{\text{78.00 g Al(OH)}_{3}} = \text{3.526 mol Al(OH)}_{3}$

3. Calculate the moles of (CH₃COO)₃Al from each reactant

$\textbf{From CH _{3} COOH:}\\\text{Moles of (CH _{3} COO) _{3} Al} = \text{2.082 mol CH _{3} COOH} \times \dfrac{\text{1 mol (CH _{3} COO) _{3} Al}}{\text{3 mol CH _{3} COOH}}\\\\= \text{0.6939 mol (CH _{3} COO) _{3} )Al}\\\textbf{From Al(OH)}_{3}:\\\text{Moles of (CH _{3} COO) _{3} Al } =\text{3.526 Al(OH)}_{3} \times \dfrac{\text{1 mol (CH _{3} COO) _{3} Al }}{\text{1 mol Al(OH)}_{3}}\\\\= \text{3.526 mol (CH _{3} COO) _{3} Al}$

$\text{Acetic acid is the \textbf{limiting reactant} because it gives fewer moles of} \\\text{(CH _{3} COO) _{3} Al. All \textbf{125 g} will react.}$