Answer to Question #183794 in General Chemistry for Lakisha Carter

Question

Na2CO3 + 2 HCl > 2 NaCl + H2O

you are working in a lab at room temperature which is considered to be 25C. The atmospheric pressure of the room from the barometer is 769 mmHg. Calculate the volume in Liters of carbon dioxide produced by the reaction of 12.78g sodium carbonate with 16.8 grams of concentrated (12.1M) hydrochloric acid.

Answer 1

The equation given above is incomplete. The correct one is:

Na₂CO₃ + 2HCl --> 2NaCl + H₂O + CO₂

First, the limiting reactant should be determined by comparing the amounts of reactants in moles. The density of concentrated HCl is not given here, but it is estimated to be approximately 1.20 g/mL. So the volume of HCl equals $\frac{16.8g}{1.20g/mL}=14.0mL=0.0140L$

$n(Na_2CO_3)=\frac{12.78g}{105.99g/mol}=0.1206mol$

$n(HCl)=12.1M\times0.0140L=0.169mol$

(0.169 : 0.1206) < (2 : 1), hence HCl is the limiting reactant.

Amount of CO₂ in moles:

$n(CO_2)=0.169mol(HCl)\times\frac{1mol(CO_2)}{2mol(HCl)}=0.0845mol$

Unit conversions:

25^oC in Kelvins is 25 + 273 = 298 K.

769 mmHg = $\frac{769}{760}=1.012atm$

Finally, solving for V according to the ideal gas law:

$V=\frac{nRT}{P}=\frac{0.0845mol\times0.08206\frac{L\cdot{atm}}{mol\cdot{K}}\times298K}{1.012atm}=2.04L$

Answer: 2.04 L