Can u rephrase this question?

Question

ok, ill give the whole question:

5.00 g of a mixture of sodium carbonate and sodium chloride was dissolved and made up to 250 mL. 25 mL of this solution was neutralized by 25 mL of .2 mol/L HCl. Find the percentage of sodium carbonate in the sample of the mixture used.

(Sodium chloride does not react with HCl)

Na2CO3 (aq) + 2HCl (aq) => 2NaCl (aq) + H2O (l) + CO2

that's the question given. I just don't understand what exactly do they want when they say "Find the percentage of sodium carbonate in the sample of the mixture used."

can u explain that to me?

thanks, after you explain then i can find the answer myself

Answer 1

5.00 g of the mixture contains X grams of Na2CO3 and (5.00-X) grams of NaCl. Find the number of grams of Na2CO3 (i.e. X) in the sample and then the percent of Na2CO3 in the 5.00 g sample  (100X/5.00 = % Na2CO3).

25 mL x 0.20 M = 5.0 millimole HCl (0.005 mole)

0.005 mole HCl will neutralize 0.0025 mole Na2CO3. Convert 0.0025 mole Na2CO3 to grams and your almost there.

Answer 2

What they mean is: what was the amount of sodium carbonate in the mixture?

Figure out the titration result, then find the percentage Na2CO3 like so.

(result/sample mass) × 100/1

The sample mass is 5 grams.