Question

Acetic Acid reacts with Sodium Carbonate to produce Sodium Acetate, Carbon Dioxide, and Water. How many milliliters of a 0.054 M Acetic Acid solution are required to completely react 100.00 g of Sodium Carbonate?

Answer 1

To react 100.00 g of sodium carbonate with a 0.054 M acetic acid solution, 34937 milliliters of the acetic acid solution are required, as determined through stoichiometry and the molarity equation.

Step-by-step explanation:

To determine how many milliliters of a 0.054 M acetic acid solution are needed to completely react with 100.00 g of sodium carbonate, we have to use stoichiometry.

First, we need the balanced chemical equation for the reaction:

2 CH3COOH (aq) + Na2CO3 (s) → 2 CH3COONa (aq) + CO2 (g) + H2O (l)

We must convert the mass of sodium carbonate to moles using its molar mass (Na2CO3: 106 g/mol):

100.00 g Na2CO3 × (1 mole Na2CO3 / 106 g Na2CO3) = 0.94340 moles Na2CO3

From the balanced equation, we see that 2 moles of acetic acid react with 1 mole of sodium carbonate.

Therefore, the moles of acetic acid needed are:

0.94340 moles Na2CO3 × (2 moles CH3COOH / 1 mole Na2CO3) = 1.88680 moles CH3COOH

To find the volume of 0.054 M acetic acid solution required, we use the molarity equation:

Volume (L) = Moles / Molarity

Volume (L) = 1.88680 moles / 0.054 M = 34.937 L × 1000 mL/L = 34937 mL

Therefore, you would need 34937 milliliters of a 0.054 M acetic acid solution to react with 100.00 g of sodium carbonate.

Answer 2

To react completely with 100.00 g of sodium carbonate, 34930 mL of a 0.054 M acetic acid solution are required, calculated using the molar mass of sodium carbonate and the stoichiometry of the reaction.

Step-by-step explanation:

To calculate the volume of a 0.054 M acetic acid solution required to react with 100.00 g of sodium carbonate, we must first write a balanced chemical equation for the reaction:

CH3COOH(aq) + Na2CO3(s) → 2 CH3COONa(aq) + CO2(g) + H2O(l)

Next, we calculate the number of moles of sodium carbonate:

Molar mass of Na2CO3 = (2×23.00) + 12.01 + (3×16.00) = 105.99 g/mol

Moles of Na2CO3 = 100.00 g / 105.99 g/mol = 0.943 mol

From the equation, 1 mole of Na2CO3 reacts with 2 moles of CH3COOH, thus 0.943 mol of Na2CO3 would require 2×0.943 mol = 1.886 mol of CH3COOH.

Now, utilizing the molarity of the acetic acid solution (M = mol/L) and rearranging for volume (L = mol/M), we find:

Volume of acetic acid = 1.886 mol / 0.054 M = 34.93 L

Convert liters to milliliters: 34.93 L × 1000 mL/L = 34930 mL

Therefore, to completely react with 100.00 g of sodium carbonate, 34930 mL of a 0.054 M acetic acid solution are required.