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6. A 1.25 g sample of impure sodium carbonate requires 18.20 mL of 0.480 M HCl to become completely neutralized. What is the percentage of the sodium carbonate in the original sample?

Question

6. A 1.25 g sample of impure sodium carbonate requires 18.20 mL of 0.480 M HCl to become completely neutralized. What is the percentage of the sodium carbonate in the original sample?

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Step 1

The mathematical expression for percentage purity :

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Step 2

The complete neutralization reaction of Sodium Carbonate and HCl can be shown through the below balanced chemical equation:

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Step 3

Given, in the above question,

Mass of impure sodium carbonate = 1.25 gm

Volume of HCl needed for complete ...

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