How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?

To determine how many milliliters of 0.1 M HCl are needed to react completely with a 1 g mixture of Na2CO3 and NaHCO3 in equimolar amounts, we first need to understand the chemical reactions involved and the stoichiometry of the reactants.

Understanding the Components

In the mixture, we have sodium carbonate (Na2CO3) and sodium bicarbonate (NaHCO3). Both of these compounds will react with hydrochloric acid (HCl) in different ways:

• Reaction with Na2CO3:

  Na2CO3 + 2 HCl → 2 NaCl + H2O + CO2↑

• Reaction with NaHCO3:

  NaHCO3 + HCl → NaCl + H2O + CO2↑

Calculating Molar Masses

Next, we need to calculate the molar masses of both compounds:

• Molar mass of Na2CO3 = 2(23.0) + 12.0 + 3(16.0) = 106.0 g/mol
• Molar mass of NaHCO3 = 23.0 + 1.0 + 12.0 + 3(16.0) = 84.0 g/mol

Finding the Amounts in the Mixture

Since the mixture contains equimolar amounts of Na2CO3 and NaHCO3, we can let x be the number of moles of each in the 1 g mixture. The total mass of the mixture can be expressed as:

106.0x + 84.0x = 1 g

This simplifies to:

190.0x = 1 g

Solving for x gives:

x = 1 g / 190.0 g/mol ≈ 0.00526 moles

Calculating Moles of HCl Required

Now, we can find out how many moles of HCl are needed for the reactions:

• For Na2CO3: 2 moles of HCl are required for every mole of Na2CO3. Therefore, for 0.00526 moles of Na2CO3, we need:

2 × 0.00526 = 0.01052 moles of HCl

• For NaHCO3: 1 mole of HCl is required for every mole of NaHCO3. Therefore, for 0.00526 moles of NaHCO3, we need:

0.00526 moles of HCl

Adding these together gives the total moles of HCl required:

Total HCl = 0.01052 + 0.00526 = 0.01578 moles of HCl

Converting Moles of HCl to Volume

Now, we can convert moles of HCl to volume using the molarity formula:

Volume (L) = Moles / Molarity

Substituting the values:

Volume = 0.01578 moles / 0.1 M = 0.1578 L

To convert this to milliliters:

Volume (mL) = 0.1578 L × 1000 mL/L = 157.8 mL

Final Result

Therefore, approximately 157.8 mL of 0.1 M HCl is required to react completely with the 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both. This calculation illustrates the importance of stoichiometry in chemical reactions and how to apply it to determine the necessary reactants for a complete reaction.