Exploring the Properties of Gases. Pre-lab questions. In an experiment carried o

Question

Exploring the Properties of Gases. Pre-lab questions. In an experiment carried out on a fixed amount of gas at constant pressure the following volume versus temperature data was collected: Use EXCEL or some other graphing program to plot this data. Carry out the appropriate extrapolation to determine the value of the temperature as the volume approaches zero. Attach your graph and the analysis to this sheet. 1. VolumeTemperature (liter) 15 12 10 458 312 214 166 A 10.0 liter vessel at 25.0° C contains 2.00 g of P (s). It is then filled with 850 torr of Ox(g) and the Pa(s) is converted to POro(s). What is the final 0:(g) pressure when all of the Pa(s) has been consumed? 2. The reduction ofacetylene, HCrCH to ethane, H3C-CH, can be carried out using a Pd catalyst: HC-CH(g) + 2H2(g) H3C-CH3(g). A 10.0 liter vessel is pressurized with 2.00 atm. of acetylene and 8.00 atm. of molecular hydrogen at 300 K. If the reaction proceeds to completeness, i.e. all of the limiting reagent is consumed, what is the final pressure? 3· 10

Explanation / Answer

2)

molar mass of P4 = 124g/mol

molar mass of O2 = 32g/mol

molar mass of P4O10 = 284g/mol

We have balanced chemical reaction equation:

P4 + 5O2 -----> P4O10

1 mol = 124g reacts 5 mol = 32x5 =160g 1mol = 284g

2g will react with = 5 mol /124g x 2g oxygen = 0.080645 mol

total moles of O2 initially present can be calculated as n = PV/RT = 850 Torr x 10L / 62.36 L.Torr.K1.mol1 x 298K

n = 0.457400 moles

moles remaining after reaction = moles present initially - moles reacted = 0.457400 - 0.080645 = 0.376755 moles

pressure (P) = nRT/V = 0.376755 mol x 62.36 L.Torr.K1.mol1 x 298K / 10L = 700.135 Torr

~= 700 Torr

3.

C2H2 + 2H2 ------->C2H6

moles of ethylene (ne) = PV/RT = 2atm x 10L/0.08206atm.L/mol.K x 300K = 0.8124 mol

moles of hydrogen (nH) = PV/RT = 8atm x 10L/(0.08206atm.L/mol.K x 300K )= 3.2496 mol

1 mol ethylene needs 2 mol hydrogen

0.8124 mol ethylene will need = 2/1 x 0.8124 = 1.6248 mol hydrogen

moles of hydrogen remaining after reaction = 3.2496 mol - 1.6248 mol = 1.62477 mol

Pressure = nRT/V = 1.62477 mol x 0.08206 atm.L/mol.K x 300K / 10.0L = 4 atm

Ans = 4 atm

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