Bob Burgess
01/10/2024 · High School
\( \left. \begin{array} { l } { H _ { 3 } C _ { 6 } H _ { 5 } O _ { 7 } + 3 NaOH \rightarrow Na _ { 3 } C _ { 6 } H _ { 5 } O _ { 7 } + 3 H _ { 2 } O } \\ { \frac { 0,1 moles NaOH } { 1 L } \times 2 mLL \times \frac { 1 L } { 1000 } = 0,002 moles NaO } \end{array} \right. \)
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La reacción es de neutralización entre ácido cítrico y hidróxido de sodio. Se calculó que en 2 mL de una solución de \( NaOH \) al 0.1 M hay 0.002 moles de \( NaOH \).
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