Answer to Question #183666 in General Chemistry for Keyla

Question

Na 2 CO 3(s) +2 HCl (aq) 2 NaCl (s) +H 2 O (g) +CO 2(g)

You are working in a laboratory at room temperature , which is considered to be 25 degrees * C 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