Chemistry

A 0.450-gram sample of impure CaCO3(s) is dissolved in 50.0 mL of 0.150 M HCl(aq). The equation for the reaction is:
CaCO3(s)+2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g)

The excess HCl(aq) is titrated by 9.75 mL of 0.125 M NaOH(aq). Calculate the mass percentage of CaCO3(s) in the sample.

Here's what I did. Is this correct? If not how do I approach it?:
excess HCl - x/.00975 L = 0.125 M
x = .00121875 mol

total HCl - x/0.05 L = 0.150 M
x = 0.0075 mol

0.0075 - 0.00121875 mol = 0.00628125 mol
0.00628125 mol HCl X 1mol CaCO3/2mol HCl X 100.0869 g/1 mol CaCO3 = 0.3143 g
0.3143 g/0.450 g = 69.8%?

  1. 👍
  2. 👎
  3. 👁
  1. That looks good to me.

    1. 👍
    2. 👎

Respond to this Question

First Name

Your Response

Similar Questions

  1. chemistry

    Calcium carbonate CaCO3 reacts with stomach acid (HCl, hydrochloric acid) according to the following equation: CaCO3(s)+2HCl(aq)->CaCl2(aq)+H2O(l)+CO2(g) Tums, an antacid, contains CaCO3. If Tums is added to 35.0 mL of 0.300 M

  2. Chemistry

    For the reaction below, Kp = 1.16 at 800.°C. CaCO3(s) CaO(s) + CO2(g) If a 25.0-g sample of CaCO3 is put into a 10.2-L container and heated to 800.°C, what percent of the CaCO3 will react to reach equilibrium?

  3. Chemistry

    1.2048g sample of impure Na2CO3 is dissolved and allowed to react with a solution of CaCl2 resulting CaCO3 after precipitation, filtaration and drying was found to weight 1.0362g. calculate the percentage purity of Na2CO3

  4. chem

    An impure sample of (COOH)2 ·2 H2O that has a mass of 3.4 g was dissolved in water and titrated with standard NaOH solution. The titration required 42.7 mL of 0.156 molar NaOH solution. Calculate the percent (COOH)2 · 2 H2O in

  1. Chemistry

    A sample of impure magnesium was analyzed by allowing it to react with excess HCl solution: Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) After 1.27 g of the impure metal was treated with 0.100 L of 0.768 M HCl, 0.0125 mol HCl remained.

  2. Analytical chemistry

    A 0.6334 gram of sample of impure mercury (II) oxide was dissolved in an unmeasured excess of potassium iodide. Reaction: HgO + 4I- + H2O ------→HgI42- + 2OH- Calculate the % HgO in the sample if titration of the liberated

  3. chemistry

    A 5.309 g antacid tablet, with CaCO3 as the active ingredient, was mixed was 30.0 mL of 0.831 M HCl. After the reaction occurred, it took 15.3 mL of 0.7034 M NaOH to neutralize the excess acid. a) How much HCl (in mL) of was

  4. Chemistry

    Determine the mass of CaCO3 in your TUMS tablet? The weight of the tablet was 0.013g Concentration of HCl was 0.1494 M Volume of HCl was 95 mL Concentration of NaOH was 0.1422 M The average amount it took to titrate was 2.13 mL I

  1. chemistry

    A 1.857 gram sample containing an unknown amount of arsenic trichloride and the rest inerts was dissolved into a NaHCO3 and HCl aqueous solution. To this solution was added 1.540 grams of KI and 50.00 mL of a 0.00912 M KIO3

  2. chemistry

    3. After removing membranes from an eggshell, the shell is dried and its mass is recorded as 5.613 g. The egg is transferred to a 250 mL beaker and dissolved in 25 mL of 6 M HCL. After filtering, the solution containing the

  3. CHEM AB HELP ASAP

    I have a pre lab questions that are due in a few hours and I need help. Theres several questions In the titration of 25.00 ml of a water sample, it took 20.590 ml of 4.995x10-3 M EDTA solution to reach the endpoint. The total

  4. College-Chemistry

    NaOH(s)+ H2SO4(aq)=Na2SO4(aq) + H2O(l) Consider the unbalanced equation above.A 0.900 g sample of impure NaOH was dissolved in water and required 37.0 mL of 0.145 M H2SO4 solution to react with the NaOH in the sample. What was the

You can view more similar questions or ask a new question.