Question

How many milliliters of a 0.8 M solution of citric acid would be needed to react with 15 grams of baking soda? citric acid (C6H8O7) baking soda (NaHCO3)the equation that demonstrates reaction:C6H8O7 + 3NaHCO3 → Na3C6H5O7 + 3H2O + 3CO2

Answer 1

74.4mL

Explanations:

Given the reaction between citric acid and baking soda expressed as:

`C_6H_8O_7+3NaHCO_3\rightarrow Na_3C_6H_5O_7+3H_2O+3CO_2`

Given the following parameters

Mass of baking soda = 15grams

Determine the moles of baking soda

`\begin{gathered} moles\text{ of baking soda}=(15)/(84.007) \\ moles\text{ of baking soda}=0.1786moles \end{gathered}`

According to stoichiometry, 1 mole of citric acid reacted with 3 moles of baking soda, the moles of citric acid required will be:

`\begin{gathered} moles\text{ of citric acid}=\frac{1mole\text{ of citric acid}}{3\cancel{moles\text{ of baking soda}}}*0.1786\cancel{moles\text{ of baking soda}} \\ moles\text{ of citric acid}=0.0595moles \end{gathered}`

Determine the volume of citric acid

`\begin{gathered} volume\text{ of citric acid}=(mole)/(molarity) \\ volume\text{ of citric acid}=(0.0595)/(0.8) \\ volume\text{ of citric acid}=0.0744L=74.4mL \end{gathered}`

Hence the required volume of citric acid is 74.4mL