chemistry

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What weight of na2co3 of 95% purity would be required to neutralize 45.6ml of 0.235n acid?

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answer 0.5978gm

  • chemistry -

    mL acid x N acid = milliequivalents acid = 45.6 mL x 0.235 = ? m.e. acid
    1 m.e. of anything = 1 m.e. of anything else; therefore, m.e. acid = m.e. Na2CO3.
    ?m.e. acid = ? m.e. Na2CO3.
    m.e. Na2CO3 x milliequivalent weight Na2CO3 = grams Na2CO3
    ?m.e. Na2CO3 x 0.053 = grams Na2CO3.
    Put all of this together to make
    mL x N x mew = grams
    45.6mL x 0.235N x 0.053g/m.e. =0.56795 g of 100% Na2CO3. Since it is only 95%, then
    0.56795/0.95 = 0.5978g which rounds to 0.598g to three significant figures.

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