Respuesta :
Answer:
74.4mL
Explanations:
Given the reaction between citric acid and baking soda expressed as:
[tex]C_6H_8O_7+3NaHCO_3\rightarrow Na_3C_6H_5O_7+3H_2O+3CO_2[/tex]Given the following parameters
Mass of baking soda = 15grams
Determine the moles of baking soda
[tex]\begin{gathered} moles\text{ of baking soda}=\frac{15}{84.007} \\ moles\text{ of baking soda}=0.1786moles \end{gathered}[/tex]According to stoichiometry, 1 mole of citric acid reacted with 3 moles of baking soda, the moles of citric acid required will be:
[tex]\begin{gathered} moles\text{ of citric acid}=\frac{1mole\text{ of citric acid}}{3\cancel{moles\text{ of baking soda}}}\times0.1786\cancel{moles\text{ of baking soda}} \\ moles\text{ of citric acid}=0.0595moles \end{gathered}[/tex]Determine the volume of citric acid
[tex]\begin{gathered} volume\text{ of citric acid}=\frac{mole}{molarity} \\ volume\text{ of citric acid}=\frac{0.0595}{0.8} \\ volume\text{ of citric acid}=0.0744L=74.4mL \end{gathered}[/tex]Hence the required volume of citric acid is 74.4mL
