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%

i think is conclusions.i think so

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol