Question

10/110 Section CHE 100 Section Experiment Date: Lab Partner: Experimental Determination of the Gas Constant Experimental Data A (a) Mass of Magnesium metal used (b) Volume of H, a collected (c) Temperature of collected () Atmospheric Pressure (c) Temperature of Water in bath (bucket) Vapor Pressure of Water above temperature 0.02659 9. 1220c Zalm Hg 2. c 12.2.. Data Analysis • Using your experimental de determine the value of the count Show all your conversions and calculations for each step clearly in the table below. Pay attention to units and significant figures Trial 1 Trial 2 Temature of in K) Moles of H as Page 1 of 2

Answer 1

Trial 1.

Volume of H₂ (V) = 27 mL = (27/1000) = 0.027 L.

Temperature of H₂ (T) = 22.0 + 273 = 295 K.

Moles of H_2(g) (n) = moles of Mg reacted

= Mass of Mg/molar mass

= (0.0265/24)

= 0.0011

Pressure of H (g) ; P = atomospheric pressure - vapor pressure of water

=( 771 - 18.7) mm Hg

= 752.3 mm Hg.

= (752.3/760) atm

= 0.989 atm.

Now, using ideal gas equation

PV = nRT

Or, R = (PV/nT) = ( 0.989×0.027)/(0.0011×295)

Or, R = 0.08228 L-atm/mol.K.

Percentage error =

(|theoretical value -calculated value|×100/theoretical value)

= (0.082057 - 0.08228)*100/0.082057

= 0.283%.

Questions.

Moles of Al = mass / molar mass of Al

= (3.00/27) = $\frac{1}{9}$

Mole ratio of Al and H₂ = 2:3

Then moles of H₂ = $\frac{2}{3}$ × moles of Al = $\frac{3}{2}$ ×$\frac{1}{9}$ = 0.167

T = 24+273 = 297 K

V = 850 mL = 0.850 L

Using ideal gas equation

PV = nRT

Or, P = (nRT/V)

Or, P = (0.167× 0.082×297/0.850) = 4.78 atm.

2)

At STP

P = 1 atm, T = 273 K.

Then, V = (nRT/P) = (0.167× 0.082× 273/1)

= 3.74 L

Volume of ballon at STP is 3.74 L.

3)

At T = 11.2 +273 = 284.2 K

P = 438 mm Hg = (438/760) = 0.576 atm.

So,

V = ( nRT/P)

= {( 0.167×0.082×284.2)/0.576}

= 6.76 L.

New volume in ballon is 6.76 L