.567L of .600M Hydrochloric acid (HCL) reacts with 5.00g of zinc metal. The gaseous product of this reaction is collected over water at 26 degrees C and 100.5 kPa. How many liters of dry gas would be collected at STP conditions?
This is a challenge problem from my chemistry class, and i can't figure it out. If someone could please help me and show me the steps to solve this problem that would be awesome! Thanks!
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Whenever you collect a gas above water, the final pressure of the gas will be due to the gas collected plus the vapor pressure of water. At 26 C, the vapor pressure of H2O = 3.4 kPa (Wikipedia). So
P H2 gas = total P - P H2O = 100.5 kPa - 3.4 kPa = 97.1 kPa.
97.1 kPa x (1 atm / 101.3 kPa) = 0.959 atm
The reaction between Zn and HCl is
Zn + 2HCl ==> ZnCl2 + H2
moles HCl = Molarity HCl x L HCl = (0.600)(0.567) = 0.340 moles HCl
5.00 g Zn x (1 mole Zn / 65.38 g Zn) = 0.0765 moles Zn
The balanced equation tells us that it takes 2 moles of HCl to react with 1 mole of Zn.
0.0765 moles Zn x (2 moles HCl / 1 mole Zn) = 0.153 moles HCl needed to react with Zn.
Obviously we have plenty of HCl (0.340 moles), so the Zn will be used up first and will determine how much product (H2) we can make.
0.0765 moles Zn x (1 mole H2 / 1 mole Zn) = 0.0765 moles H2 produced.
So we now have moles of H2, pressure of H2, the temperature (36 C = 309 K), and R. Use the ideal gas law to calculate the volume of H2 gas.
PV = nRT
V = nRT / P = (0.0765 moles)(0.0821 L atm / K mole)(309 K) / (0.959 atm) = 2.02 L H2 gas