How many milliliters of 0.1 m hcl is required to react with 0.15g sodium carbonate (na2co3?

1) Write the balanced chemical equation

     2HCl + Na2 CO3 > 2NaCl + H2CO3

2) Write the molar ratios:

    2 mol HCl : 1 mol Na2CO3 : 2 mol NaCl : 1 mol H2CO3

3) Convert 0.15g of sodium carbonate to number of moles

3a) Calculate the molar mass of Na2CO3

Na: 2 * 23 g/mol = 46 g/mol

C: 12 g/mol =

O: 3 * 16 g/mol = 48 g/mol

molar mass = 46g/mol + 12g/mol + 48g/mol = 106 g/mol

3b.- Calculate the number of moles of Na2CO3

# moles = grams / molar mass = 0.15 g / 106 g/mol = 0.0014 mol Na2CO3

4) Calculate the number of moles of HCl from the molar proportion:

[0.0014 mol Na2CO3] * [2 mol HCl / 1 mol Na2CO3] = 0.0028 mol HCl

5) Calculate the volume of HCl from the definition of Molarity

Molarity, M = # moles / volume in liters

=> Volume in liters = # moles / M = 0.0028 mol / 0.1 M = 0.028 liters

0.028 liters * 1000 ml / liter = 28 ml.

28 mililiters of 0.1 M HCl.

Ireally don’t know this