The percent by mass of bicarbonate (HCO3-) in an Alka-Seltzer tablet is 32.5 percent. Calculate the volume (in mL) of CO2 generated at 37.0^C and 0.900 atm when a person ingests a 3.29 gram tablet.
(Hint: A reaction occurs between the HCO3- ions and HCl in the stomach.)
(Hint: A reaction occurs between the HCO3- ions and HCl in the stomach.)
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first work out the mass of HCO3- in thart mss of tablet
32.5 / 100 x 3.29 g = 1.06925 g
moles HCO3- = mass / molar mass
= 1.06925 g / 61.018 g/mol
= 0.01752 mol
write a balanced eqaution
HCO3^- + HCl ----> Cl- + CO2(g) + H2O
1 mole of CO2 forms from 1 mole of HCO3^-
Thus moles CO2 = moles HCO3^-
= 0.01752 mol CO2
Now, use the ideal gas eqaution to work out what volume this will have under the given conditions
PV = nRT
P = 0.900 atm
V = ? L
n = 0.01752 mol
R = gas constant = 0.082057 Latmmol^-1K^-1
T = temp in Kelvin = 37.0 deg C + 273.15 = 310.15 K
V = nRT / P
= 0.01752 mol x 0.082057 x 310.15 K / 0.900 atm
= 0.495 L
= 495 ml (3 sig figs)
32.5 / 100 x 3.29 g = 1.06925 g
moles HCO3- = mass / molar mass
= 1.06925 g / 61.018 g/mol
= 0.01752 mol
write a balanced eqaution
HCO3^- + HCl ----> Cl- + CO2(g) + H2O
1 mole of CO2 forms from 1 mole of HCO3^-
Thus moles CO2 = moles HCO3^-
= 0.01752 mol CO2
Now, use the ideal gas eqaution to work out what volume this will have under the given conditions
PV = nRT
P = 0.900 atm
V = ? L
n = 0.01752 mol
R = gas constant = 0.082057 Latmmol^-1K^-1
T = temp in Kelvin = 37.0 deg C + 273.15 = 310.15 K
V = nRT / P
= 0.01752 mol x 0.082057 x 310.15 K / 0.900 atm
= 0.495 L
= 495 ml (3 sig figs)