In the reaction between sodium carbonate and hydrochloric acid, carbon dioxide i
ID: 879228 • Letter: I
Question
In the reaction between sodium carbonate and hydrochloric acid, carbon dioxide is produced via the following stoichiometry.
2 HCl (aq) + Na2CO3 (s) -> CO2 (g) + 2 NaCl (aq) + H2O (l)
1) If 50.0 g of Na2CO3 is used to neutralize 50 mL of 0.1 M HCl, how many moles of CO2 will be produced?
2) If all of the CO2 gas is captured in 250 mL container, what will be the pressure of the CO2 inside the vessel at 25 C?
3) What would be the volume of the same sample of gas at STP?
4) What would be the pressure inside the 250 mL container at 25C if all of the Na2CO3 was reacted away? (5 points)
Explanation / Answer
2HCl(aq) + Na2CO3(s) ----->2NaCl (aq) + CO2 (g) + H2O (l)
1-MwNa2CO3=106g/mol
50 g of Na2CO3 = 0.472mole of Na2CO3
mole of HCl = 50x0.1/1000 = 5x10^-3 mole
2mole of HCl react with 1mole of Na2CO3
5x10^-3 mole HCl react with 2.5x10^-3 mole of Na2CO3 &
produce 2.5x10^-3 mole CO2
2--n= 2.5x10^-3 V=0.25l T = 273+25= 298K
PV= nRT
P= nRT /V= 2.5x10^-3 x0.082 x298 / 0.25 = 0.244 atm
3- V= nRT/P
at STP
P= 1atm T =273K
V= 2.5x10^-3 x 0.082 x 273 /1 = 5.59x10^-2 L
4-mole of CO2 formed when all Na2CO3 reacted = 0.472 mole
P= nRT / V
P = 0.472 x 0.082 x 298 / .25 = 46.135 atm