# Question #05bc4

May 27, 2016

You need 2.6 g of baking soda or 35 mL of vinegar, whichever is less.

#### Explanation:

The equation for the reaction is

$\text{NaHCO"_3 + "HC"_2"H"_3"O"_2 → "NaC"_2"H"_3"O"_2 + "CO"_2 + "H"_2"O}$

Step 1. Calculate the moles of sodium acetate

${\text{Moles of NaC"_2"H"_3"O"_2 = 2.5 color(red)(cancel(color(black)("g NaC"_2"H"_3"O"_2))) × ("1 mol NaC"_2"H"_3"O"_2)/(82.03 color(red)(cancel(color(black)("g NaC"_2"H"_3"O"_2)))) = "0.0305 mol NaC"_2"H"_3"O}}_{2}$

Step 2. Calculate the mass of baking soda required

${\text{Moles of NaHCO"_3 = 0.0305 color(red)(cancel(color(black)("mol NaC"_2"H"_3"O"_2))) × ("1 mol NaHCO"_3)/(1 color(red)(cancel(color(black)("mol NaC"_2"H"_3"O"_2)))) = "0.0305 mol NaHCO}}_{3}$

${\text{Mass of NaHCO"_3 = 0.0305 color(red)(cancel(color(black)("mol NaHCO"_3))) × ("84.01 g NaHCO"_3)/(1 color(red)(cancel(color(black)("mol NaHCO"_3)))) = "2.6 g NaHCO}}_{3}$

Step 3. Calculate the volume of acetic acid

${\text{Moles of HC"_2"H"_3"O"_2 = 0.0305 color(red)(cancel(color(black)("mol NaC"_2"H"_3"O"_2))) × ("1 mol HC"_2"H"_3"O"_2)/(1 color(red)(cancel(color(black)("mol NaC"_2"H"_3"O"_2)))) = "0.0305 mol HC"_2"H"_3"O}}_{2}$

${\text{Volume of HC"_2"H"_3"O"_2 = 0.0305 color(red)(cancel(color(black)("mol HC"_2"H"_3"O"_2))) × ("1 L HC"_2"H"_3"O"_2)/(0.87 color(red)(cancel(color(black)("mol HC"_2"H"_3"O"_2)))) = "0.035 L HC"_2"H"_3"O"_2 = "35 mL HC"_2"H"_3"O}}_{2}$

You need either ${\text{2.6 g of NaHCO}}_{3}$ or ${\text{35 mL of HC"_2"H"_3"O}}_{2}$, whichever is less (the limiting reactant).