Práctica 9 benzoato de etilo
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CÁLCULOS
Cálculos estequiométricos
C2H5OH
m C2H5OH = V ∗ ρ = 50mL ∗ 0.816 g = 40.8 g
mL
n C2H50H =
C7H6O2
m C7H602 = 10g
n C7H602 =
H2SO4
40.8 g
46.07 g = 0.8856 mol
mol
10g
= 0.08188 mol
122.13 g/mol
m H2SO4 = 1 ml ∗ 1.841 g = 1.841 g
ml
n H2SO4 =
1.841 g
98.08 g = 0.01877 mol
mol
“Acido benzoico es el reactivo limitante”
• Resultados de la experimentación
Peso del producto a obtener
m C9H10O2 = m C7H602 ∗
m C9H10O2 = 10 g ∗ 150.17 g
mol
122.13 g = 12.2959 g
mol
PM C9H10O2
PM C7H602