To calculate the percentage by mass of sodium chloride in the mixture, we need to determine the amount of sodium chloride present in the 25 mL of the solution that required 21 mL of 0.1 M hydrochloric acid for neutralization.
First, let's calculate the number of moles of hydrochloric acid used in the neutralization reaction:
Moles of hydrochloric acid = concentration of hydrochloric acid × volume of hydrochloric acid Moles of hydrochloric acid = 0.1 mol/L × 0.021 L Moles of hydrochloric acid = 0.0021 mol
Since the stoichiometry of the neutralization reaction between hydrochloric acid (HCl) and sodium chloride (NaCl) is 1:1, the moles of hydrochloric acid used is equal to the moles of sodium chloride in the mixture.
Now, let's calculate the moles of sodium chloride:
Moles of sodium chloride = 0.0021 mol
Next, we need to determine the total mass of the mixture. The mixture contains both sodium carbonate (Na2CO3) and sodium chloride (NaCl). Let's assume the entire 2.5 g mixture is used to make the 250 mL solution.
Total mass of the mixture = 2.5 g
Now, let's calculate the percentage by mass of sodium chloride:
Percentage by mass of sodium chloride = (Moles of sodium chloride × molar mass of sodium chloride) / Total mass of the mixture × 100
The molar mass of sodium chloride (NaCl) is approximately 58.44 g/mol.
Percentage by mass of sodium chloride = (0.0021 mol × 58.44 g/mol) / 2.5 g × 100
Percentage by mass of sodium chloride ≈ 9.82%
Therefore, the percentage by mass of sodium chloride in the mixture is approximately 9.82%.