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kl 發問於 科學及數學化學 · 1 十年前

f.5 molar volume calculations

An antacid tablet containing sodium hydrogencarbonate completely reacted with 2.0M hydrochloric acid. 0.420g of the tablet produced 114cm^3 of carbon dioxide, measured at room temperature and pressure.What is the percentage by mass of sodium hydrogencarbonate in the tablet?

molar volume=24.0dm^3mol-1

2 個解答

  • 1 十年前

    no. of mole of CO2 = (114/1000)/24

    = 0.004875 mol

    NaHCO3 + HCl → NaCl + H2O + CO2

    mole ratio of CO2 : NaHCO3 = 1 : 1

    so, no. of mole of NaHCO3 = 0.004875 mol

    mass of NaHCO3 = 0.004875 x 84

    = 0.4095 g

    percentage by mass = (0.4095-0.420)x100%

    = 97.5%

    concentration of hydrochloric acid is not used. it has no use, just to distract you. it is only useful if volume of acid is asked.

    2006-10-31 23:12:43 補充:

    ~~sorry~~no. of mole of CO2 = 0.00475 mol, as well as no. of mole of NaHCO3.mass of NaHCO3 = 0.399 gpercentage by mass = (0.399/0.420)x100%= 95%~~SORRY~~

  • 1 十年前

    What is the ans. of this question ? PLEASE CHECK

    To me. i think the ans should be 90%

    as the equation can be resprensented in this way:

    NaHCO3+HCl -- H2O+CO2+NaCl (balanced)

    by the information of 114 cm3 of CO2

    we can know that there are 0.00475mole of CO2 produced

    coz (114/1000)/24 =0.0045 mole

    thus the no. of mole of NaHCO3 = 0.0045mole (ratio by the equation)

    Hence , find the mass of pure NaHCO3 = 0.0045X(23+1+12+16X3)

    = 0.0045 X 84 =0.378g

    thus the pencetage is

    0.378/0.42 X 100% = 90%

    I suspect whether i am true or not...because i haven't use the information - 2M HCl

    資料來源: ME