Suppose 1.02 g of Zn metal was added to 20.0 mL of 3.00 M HCl in a closed 50.0 mL vessel in 298 K and 0.000 atm of pressure.
What is the final pressure in the vessel?
If initially there was 1.00 atm of air present in the closed 50.0 mL vessel, what would be the final pressure?
volume of the free space = (50.0 - 20.0) = 30.0 ml = 0.030 L
1.02 g of Zn = mass / molar mass = 1.02 g / 65.38 g / mole = 0.0156 mole
20.0 mL of 3.00 M HCl = 0.020 L * 3.00 mole / L = 0.06 mole.
reaction is
Zn + 2 HCl ...............> ZnCl2 + H2 (g)
one mole Zn react with 2 mole HCl to produce one mole H2 gas and one mole ZnCl2.
Zn is the limiting reactant.
thus
mole of H2 produce = 0.0156 mole.
ideal gas equation is
PV = nRT
P = nRT / V = (0.0156 mole * 0.0821 L-atm / mole-K * 298 K) / 0.030 L = 12.72 atm
final pressure in the vessel = 12.72 atm
initially there was 1.00 atm of air present in the closed 50.0 mL vessel:
then
final pressure = (1.00 + 12.72) = 13.72 atm