How do you determine the volume of hydrogen collected at STP?
it was a reaction of magnesium with hydrochloric acid:
Mg(s) + 2HCl(aq) ---> MgCl2(aq) + H2(g)
values you may need:
Volume of hydrogen gas (mL) = 48.3 ml
STP = 22.4 L/mol
room temperature = 22 degrees Celsius
barometric pressure = 100.5 kPa
water vapor pressure at 22 degrees Celsius = 2.60 kPa
Mass of Mg Ribbon (g) = 0.0500 g
- pisgahchemistLv 79 years agoFavorite Answer
1 mole of any gas at STP has a volume of 22.4L.
Clearly, the H2 gas you collected was not at STP. It would seem that you collected 48.3 mL at 22C and 100.5 kPa. Use the combined gas law to find the volume at STP.
Orig is P1, V1 and T1 ... V2 is the volume at STP
P1V1/T1 = P2V2/T2
V2 = P1V1T2/(T1P2)
V2 = (100.5 - 2.60)kPa x 48.3 mL x 273K / 295K / 101.3 kPa = 43.2 mL
Subtract the water vapor pressure to get the initial pressure of H2 gas.
=========== Follow up ===============
OD92 didn't take into account the vapor pressure of water due to collecting H2 over water. Also, it's rather odd to be measuring a volume in cubic meters when the original volume was in mL.
- Anonymous9 years ago
Im assuming your looking for the method?
First, using the mass of Mg solid used in the reaction, you calculate the moles of Mg used, using the equation: moles = mass/ relative molecular mass(RMM) (RMM Mg = 24.305 g/mol).
You should get 2.06EXP-3 (0.00206) moles.
Now from the chemical equation, you know that the mole ratio between Mg and H2 is 1:1.
Therefore the number of moles of Mg used is equal to the number of moles of H2 produced.
Using that value for the number of moles of H2, use the equation PV = nRT.
P = pressure = 100.5 kPa (or 101325 Pa at STP)
V = volume = ? m^3
n = number of moles of H2 gas
R = universal gas constant = 8.314 J/K/mol
T = absolute temperature = 295.15 K (or 273.15 K at STP)
You should get 0.000046 m^3 of H2 gas produced, which when converted is almost the same answer as you have.
(N.B. There is an assumption here however, that the hydrogen gas is behaving like an ideal gas).