Chemistry lab problem.?
So I am doing STP calculations.
I found the STP of the gas collected. Trial 1 is 0.03508333 L and Trial 2 is 0.0360598 L
I am having trouble with finding molar volume. I know this because my percent error is very off.
The givens are:
Volume of hydrogen gas collected in L
Trial 1) 0.0365 L
Trial 2) 0.0374 L
Barometric pressure (kPa)
Trial 1) 99.3
Trial 2) 99.6
Mass of magnesium ribbon (grams)
Trial 1) 0.0342g
Trial 2) 0.0350 g
The lab has told me to subtract both of the bottom, the temperature of water and vapor pressure.
Temperature of water (celcius)
Trial 1) 21.2 C
Trial 2# 21.2 C
Vapour pressure of water (kPa)
trial 1) 2.5
Trial 2) 2.5
- pisgahchemistLv 73 years agoFavorite Answer
STP is "standard temperature and pressure", or 1.00 atm and 0C (273.15K). One mole of any ideal gas at STP has a volume of 22.4L
Mg(s) + 2HCl(aq) --> MgCl2(aq) + H2(g)
0.0342g ................. ........................?mol
0.0342g Mg x (1 mol Mg / 24.305g Mg) x (1 mol H2 / 1 mol Mg) = 0.001407 mol H2
Convert volume of H2 to STP using the combined gas law
P1V1 / T1 = P2V2 / T2
V2 = P1V1T2 / (T1P2)
V2 = (99.3 - 2.5) kPa x 0.0365L x 273.15K / 294.35K / 101.3 kPa
V2 = 0.03237L
Molar volume (in L/mol) = 0.03237L / 0.001407 mol H2 = 23.01 L/mol
1 mol of H2 occupies a volume of 23.01L at STP according to your data. Now repeat with the info from your second trial.