Chemistry lab problem.?

Question

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

Answer 1

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STP .....

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.