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.

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  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
    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 %

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  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

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