Chemistry Damon

1g of mixture of na2c03 and k2c03 was made upto 250ml is aqueous solution . 25ml of this solution was neutralized 20ml of hcl of unknow concentration. The neutralized solution required 16.24ml of 0.1N AGN03 for precipitation. Calculate a) the k2c03 is mixture b) conc of hcl in g/L C) molarity of hcl
A is answer 60%
B is answer 2.9g/L
c) is answer 0.0812M

Damon wants you step by step for me.

  1. 👍 0
  2. 👎 1
  3. 👁 320
  1. Na2CO3 =2(23)+12+3(16) = 106 g/mol
    x moles of Na2 CO3 mass = 106 x grams

    K2CO3 = 2(39)+12+3(16) = 138 g/mol
    y moles of K2CO3 mass = 138 y grams

    106 x + 138 y = 1 gram total

    in the titration I used 25 mL of my 250 ml so I used 1/10 gram solids of my original carbonates mixture

    The reactions are
    Na2CO3 + 2HCl --> 2NaCl + H2O + CO2
    K2CO3 + 2HCl --> 2 KCl + H2O + CO2

    Not being a chemist I will assume that I get the same number of moles of NaCl plus KCl as I had moles of AgNO3
    How many is that?
    moles AgNO3 = .1moles/liter * .01624 L = .001624 moles of NO3-
    so I had .001624 moles of NaCl plus KCl
    which means I had .000812 moles of
    Na2CO3 plus K2CO3
    x/10 moles Na2CO3 + y/10moles K2NO3 =.000812 moles
    or
    x + y = .00812
    and from way long ago
    106 x + 138 y = 1 gram total
    ---------------------
    106 (.00812-y) + 138 y = 1

    .861 -106 y + 138 y = 1
    32 y = .139
    y = .00435
    fraction of y which is K2CO3 = .00435/.00812 = .536 which is 53.6 %

    1. 👍 0
    2. 👎 0
    posted by Damon
  2. Wait a minute, x and y are moles and you want percentage of grams perhaps.
    x = .00377
    108 x = .40716
    138 y = .6003

    so yes, 60% K2CO3

    1. 👍 0
    2. 👎 0
    posted by Damon

Respond to this Question

First Name

Your Response

Similar Questions

  1. chemistry Who clearlyexplain step by step

    1g of mixture na2c03 and k2c03 was made upto 250ml is aqueous solution. 25ml of this solution was neutralized 20ml of hcl of unknown concentration. The neutralized solution requires 16.24ml of 0.1N agn03 for precipitation

    asked by Ferda on January 18, 2014
  2. chemistry Damon helps me step by step

    1g of mixture na2c03 and k2c03 was mole made upto 250ml is aqueous solution. 25ml of this solution was neutralized. 20ml of hcl of unknown concentration. The neutralized solution required 16.24ml of 0.1N AGNo3 for precipitation,

    asked by Ken Hung on January 12, 2014
  3. Chemistry

    25g of a sample of ferrous sulphate was dissolved in water containing dilute h2s04 and the volume made up to one litre. 25ml of this solution required 20ml of N/10 kmn04 solution for complete oxidation. Calculate the percentage of

    asked by Ken Hung on January 11, 2014
  4. chemistry who step by step for me

    1.2g mixture of na2c03 and k2c03 was dissolved in water to form 100cm3 of solution, 20 cm3 of this solution required 40cm3 of 0.1 hcl for neutralization. Calculate the weight na2c03 and k2c03 in the mixture? My calculation

    asked by ken on June 22, 2014
  5. chemistry (Pls Bob check for me)

    A 1.2g mixture of na2c03 and k2c03 was dissolved in water to form 100cm3 of s solution, 20cm3 of this solution required 40cm3 of 0.1N hcl for neutralization. Calculate the wieght of na2c03 and k2c03 in the mixture The answer 0.4G

    asked by Fai on February 24, 2013
  6. Chemistry

    A 0.512g sample of CaCO3 is dissolved in 12M HCl and the mixture is diluted to 250ml. A. Calculate the concentration of Ca+2 in the 250ml solution? B. How many moles of Ca+2 are in a 25ml sample of th 250ml solution?

    asked by Felix Dagogo on April 26, 2019
  7. chemistry

    25ml of a solution of na2c03 having a specific gravity of 1.25g ml required 32.9ml of a solution of hcl containing 109.5g of the acid per litre for complete neutralization. Calculate the volume of 0.84N heso4 that will be

    asked by Fai on February 17, 2013
  8. chemistry

    0.50g of a mixture of K2c03 and Li2c03 requires 30ml of a 0.25N hcl solution for neutralization. What is the percentage composition of the mixture? The answer k2c03 96% Li2c03 4% K2c03 weight 138g/mol 138/2 = 69 Li2c03 weight

    asked by Fai on February 22, 2013
  9. chenistry

    25ml of a solution of na2c03 having a sepcific gravity of 1.25g solution of hcl containing 109.5g of the acid per litre for complete neutralization. Calculate the volume of 0.84N h2s04 that will be compltetly neutralized by 125g

    asked by Fai on February 17, 2013
  10. chemistry

    0.50g of mixture of k2c03 and li2c03 requires 30ml of a 0.25N HCL solution for neutralization. What is the percentage composition of the mixture? The answer k2c03 96% The answer li2c03 4% I want Bob step by step explain. I do not

    asked by Fai on February 24, 2013

More Similar Questions