posted by .

A 0.2528 g sample of a mixture containing sodium carbonate and sand requires 12.32 mL of 0.2409 M HCl(aq). Assuming that the sand does not react with the HCl(aq), what percentage of the mixture is sodium carbonate?

Balanced equation:
Na2CO3(s) + 2 HCl(aq) ¨ 2 NaCl(aq) + CO2(g) + H2O(l)

  • chemistry -

    moles HCl = M x L = ??
    moles Na2CO3 = 1/2 moles HCl (from the equation).
    g Na2CO3 = moles x molar mass

    %Na2CO3 = (mass Na2CO3/mass sample)*100 = ??

  • chemistry -

    Thank you! So, is this correct?

    moles HCl = M x L
    = .2409 X .01232 = .00297

    moles Na2CO3 = 1/2 moles HCl (from the equation).
    = .00297 / 2 = .00148

    g Na2CO3 = moles x molar mass
    .00148 X 106 = .157

    %Na2CO3 = (mass Na2CO3/mass sample)*100 = ??
    .157 / .2528 X 100 = 60.8%

  • chemistry -

    Almost but not quite.
    1. When multiplying 0.2409 x 0.01232, the answer is 0.00296788. There are two things wrong here. First, you threw away at least one significant figure (both numbers you multiplied have four figures so you are allowed four in the answer but you have only three). I leave all of that in the calculator and use it as is in the next calculation. You COULD round at this point to 0.002968 but I recommend you leave all of the numbers in the calculator and go to the next step.
    You need to go through and correct the other steps.
    2. I think you punched in the wrong numbers for the last step because
    (0.157/0.2528)*100 = = 62.10%.

    If we go through it all at once as follows, I get this
    (0.2409 x 0.01232 x 1/2 x 105.99/0.2528)*100 = 62.21647 which rounds to 62.22% to four significant figures.

  • chemistry -

    Oh ok, I see what I did now. Thank you for the help! : )

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. chem

    "The percentage of sodium hydrogen carbonate, NaHCO3, in a powder for stomach upsets is found by titrating .275 M hydrochloric acid. if 15.5 mL of hydrochloric acid is required to react with .500 g of the sample, what is the percentage …
  2. Chemistry

    2.500g sample of a mixture of sodium carbonate and sodium chloride is dissolved in 25.00ml of 0.798M HCl. Some acid remains after the treatment of the sample. If 28.7 ml of 0.0108M NaOH were required to titrate the excess hydrochloric …
  3. Chemistry 1

    Sodium carbonate reacts with a solution of HCl solution with the evolution of gas (CO2) to yield a solution that is neutral. If 0.265 grams of sodium carbonate are required to neutralize 1.00 mL of HCl solution, what mass (in grams) …
  4. Chemistry

    What are the balanced equation for the following reactions: HCl and Sodium Hydrogen Carbonate HCl and Copper oxide HCl and Calcium carbonate HCl and Magnesium NaOH and Ammonium Chloride
  5. science

    if i had mixture of sodium carbonate and sand, and spilled some sodium carbonate (no sand spilled) after weighing the mixture, would the spill affect the percentage of sand or sodium carbonate when we calculate it?
  6. Chem

    Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 3.30 g of sodium carbonate is mixed with one containing 3.00 g of silver nitrate. How many …
  7. chemistry Please who helps me stpe by step

    2gm of mixture hydrated sodium carbonate na2c03 /OH20 and sodium bicarbonate was dissolved in water and made up to 250cc. 25cc of this solution was titrated, using methyl orange as indicator and 22.5cc of 0.087N hcl were required for …
  8. Chem

    A mixture of 5.00 g of sodium carbonate, Na2CO3 and sodium hydrogen carbonate, NaHCO3 is heated. The loss in mass is 0.31 g. Sodium carbonate does not decompose on heating. Calculate the percentage by mass of sodium carbonate in the …
  9. Chemistry

    Explain how you would obtain solid sodium carbonate from a mixture of lead carbonate and sodium carbonate powder
  10. Chemistry

    FB1 is a solution containing a mixture of potassium carbonate and sodium hydrogen carbonate . 25cm^3 of this mixture was to titrated against 0.250mol/dm^3 solution of HCL. In one titration using phenolphthalein as indicator, 10.00cm^3 …

More Similar Questions