How many milliliters of 0.1M HCl are required to completely react with a 1g mixture of Na₂CO₃ and NaHCO₃, given that both compounds are present in equimolar amounts?

To determine how many mL of 0.1 M HCl are required to react completely with a 1g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both, we need to first calculate the molar mass of each compound and then use stoichiometry to find the amount of HCl required.

Calculate the molar mass of Na2CO3:
Na (Sodium) has a molar mass of approximately 22.99 g/mol.
C (Carbon) has a molar mass of approximately 12.01 g/mol.
O (Oxygen) has a molar mass of approximately 16.00 g/mol.
Molar mass of Na2CO3:
2(Na) + 1(C) + 3(O) = 2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 45.98 g/mol + 12.01 g/mol + 48.00 g/mol = 105.99 g/mol

Calculate the molar mass of NaHCO3:
Na (Sodium) has a molar mass of approximately 22.99 g/mol.
H (Hydrogen) has a molar mass of approximately 1.01 g/mol.
C (Carbon) has a molar mass of approximately 12.01 g/mol.
O (Oxygen) has a molar mass of approximately 16.00 g/mol.
Molar mass of NaHCO3:
1(Na) + 1(H) + 1(C) + 3(O) = 22.99 g/mol + 1.01 g/mol + 12.01 g/mol + 3(16.00 g/mol) = 22.99 g/mol + 1.01 g/mol + 12.01 g/mol + 48.00 g/mol = 84.01 g/mol

Since the mixture contains equimolar amounts of Na2CO3 and NaHCO3, their molar masses are approximately 105.99 g/mol and 84.01 g/mol, respectively.

Now, let's calculate the number of moles of each compound in the 1g mixture:

Moles of Na2CO3 = (mass of Na2CO3) / (molar mass of Na2CO3)
Moles of Na2CO3 = 1g / 105.99 g/mol ≈ 0.00943 moles

Moles of NaHCO3 = (mass of NaHCO3) / (molar mass of NaHCO3)
Moles of NaHCO3 = 1g / 84.01 g/mol ≈ 0.01190 moles

Now, let's find the balanced chemical equation for the reaction between Na2CO3 and HCl:

Na2CO3 + 2 HCl -> 2 NaCl + H2O + CO2

From the balanced equation, we see that 1 mole of Na2CO3 reacts with 2 moles of HCl.

So, for the moles of Na2CO3 we calculated (0.00943 moles), it will require 2 * 0.00943 moles of HCl to react completely.

Total moles of HCl required = 2 * 0.00943 moles ≈ 0.01886 moles

Now, we can calculate the volume of 0.1 M HCl required using the formula:

Volume (in liters) = Moles / Molarity

Volume (in liters) = 0.01886 moles / 0.1 moles/L = 0.1886 L

To express this volume in milliliters (mL), multiply by 1000:

Volume = 0.1886 L * 1000 mL/L = 188.6 mL

So, approximately 188.6 mL of 0.1 M HCl are required to react completely with the 1g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both.