0.406 g of antacid, active ingredient Ca(HCO3)2 treated with excess HCl gives 86.44 mL of CO2 at 25.00oC, 1.000 atm. What % is active ingredient? Asumme a perfect system (i.e 100% yield)

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We can solve this problem by employing the ideal gas law. 

PV = nRT 

where 

P = pressure (atm) = 1.000 atm

V = volume (L) = (86.44mL/1000mL)L

n = number of moles = ?

R = gas constant, 0.08206 atm-L/mol/K

T = temperature (in Kelvin) = 25.00 + 273.15 = 298.15K

 

n = PV/RT = (1.00*0.08644)/(0.08206*298.15)

   = 0.003522 moles CO2

Ca(HCO3)2 ---> Ca(OH)2 + 2CO2

There are 2 moles of CO2 per one mole of Ca(HCO3)2 so,

0.003522 moles CO2 x ( 1 mole Ca(HCO3)2 / 2 moles CO2)

        = 0.0017665moles Ca(HCO3)2

 

remember that n =  mass/ molar mass

so,  mass Ca(HCO3)2= n* molar mass

       mass Ca(HCO3)2 = 0.0017665 * 162.11464 g/mol

                                  = 0.28638 g Ca(HCO3)2

 

percent active ingredient = 0.28638/0.406 x100

                                     = 70.5% active ingredient


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