Question

Please provide the formulas or steps to solve the four short problems considering the following data:

DATA

Atmospheric Pressure: 752mmHg

Temperature of H2O: 21 deg. C

Vapor Pressure of H2O at this temp: 18.7torr

Pressure of CO2 in system: 733mmHg

Balanced reaction: NaHCO3(aq) + HCl(aq) --> CO2(g) + H2O(l)

Mass of Sample: 0.2008g

Vol initial: 16.60mL

Vol final: 35.10mL

Vol total: 20.50mL

PROBLEMS

1.) Mols of CO2

2.) Moles of NaHCO3

3.) Mass of NaHCO3 in sample (in grams)

4.) Mass percent of NaHCO3 in sample (percentage)

Answer 1

Pressure of dry gas (P) = atmospheric pressure - vapor pressure of water at the given temperature

= (752 - 18.7) torr

= 733.3 torr

= 733.3 (torr) × $\frac{1 (atm)}{760 (torr)}$

= 0.965 atm.

Volume of CO₂ (V) = 20.50 mL = 20.50 (mL) × $\frac{1 (L)}{1000 (mL)}$

= 0.0205 L.

Temperature (T) = 21 +273= 294 K

R = 0.082 L-atm/mol.K

Now , using ideal gas equation.

PV = nRT

Or, number of moles of CO₂ (n) = (PV/RT)

= (0.965×0.0205/0.082×294)

= 0.00082

Mole ratio of NaHCO₃ and CO₂ = 1:1.

So, moles of NaHCO₃ = 0.00082

Mass of NaHCO₃ in sample = moles × molar mass

= 0.00082× 84

= 0.068

Percent of NaHCO₃

= {(Mass of NaHCO₃ ×100)/mass of sample}

= (0.0688×100/0.2008)

= 34.30 %.