A 0.7860-g sample of impure solid Na2CO3 is neutralized by 23.48 mL of 0.1082 M HCl. What percentage purity of Na2CO3 was contained in the sample? There are no acidic or basic impurities present in the sample

Percentage purity of in sample is 17.13%

Explanation:

Net Ionic equation:

So, two moles of neutralize 1 mol of  

Alternatively, two moles of HCl neutralize 1 mol of

Number of moles of HCl added =

So, 0.002541 moles of HCl neutralizes 0.0012705 moles of

Molar mass of = 105.99 g/mol

So, mass of present in impure sample =

So percentage purity of in sample = % = 17.13%

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