Question

Baking soda (NaHCO3) can be added to a fruit mix solution to create a carbonated drink. An example is the reaction between baking soda and citric acid below.

C6H8O7 + 3NaHCO3 → Na3C6H5O7 + 3H2O + 3CO2

a. What is the molar mass of citric acid (C6H8O7) and baking soda (NaHCO3)?
b. How many milliliters of a 0.8 M solution of citric acid would be needed to react with 15 grams of baking soda?

I already have a done I just need b.

Answer 1

74.4 mL of a 0.8 M solution of citric acid would be needed to react with 15 grams of baking soda.

Equation that demonstrates reaction;

C6H8O7 + 3NaHCO3 → Na3C6H5O7 + 3H2O + 3CO2

Molar mass of citric acid (C6H8O7) =  12×6 + 1.008×8 + 16×7

                                                           =  192.064 g/mol

Molar mass of baking soda (NaHCO3) = 23 + 1 + 12 + 3×16

                                                                = 84 g/mol

$^c C_6H_8O_7 = 0.8 M = 0.8 mol/L\\^m NaHCO_3 =15g$

From the reaction equation:

$3$ × $^n C_6H_8O_7 = ^n NaHCO_3$

Molar mass of baking soda (NaHCO3) = 84 g/mol

So,

$^n NaHCO_3 = ^m NaHCO_3/ ^MNaHCO_3$

And;

$^n C_6H_8O_7 = ^nNaHCO_3/ 3$

From the definition of Molarity,

$^V C_6H_8O_7 =\frac{^n C_6H_8O_7}{^cC_6H_8O_7 } =\frac{^n NaHCO_3}{^3*n^cC_6H_8O_7 } = \frac{^m NaHCO_3 / ^M NaHCO^3}{3*^cC_6H_8O_7 }$

Calculating;

$^V C_6H_8O_7 = \frac{15/84}{3*0.8}$

               $= 0.0744L\\= 74.4 mL$

Therefore, 74.4 mL of a 0.8 M solution of citric acid would be needed to react with 15 grams of baking soda.

Learn more  about molarity here:

https://brainly.com/question/14469428

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