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f.5 molar volume calculations

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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

最佳解答:

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~~

其他解答:

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 HCl5C926699F268FE02