Part A Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unkno

Question

Part A

Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unknown amount of propane, C3H8) were added to the same 10.0-Lcontainer. At 23.0 C, the total pressure in the container is 5.00 atm . Calculate the partial pressure of each gas in the container.

Express the pressure values numerically in atmospheres, separated by commas. Enter the partial pressure of methane first, then ethane, then propane.

??? = atm  

Part B

A gaseous mixture of O2 and N2 contains 36.8 % nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 685 mmHg ?

Express you answer numerically in millimeters of mercury.

Poxygen= ??? mmHg

Part C

What is the pressure (in atmospheres) of the gas inside the container connected to an open-end, mercury-filled manometer as shown in the picture? (Figure 1) The atmospheric pressure is 0.95 atm.

figure 1

Express your answer numerically in atmospheres.

gas pressure = ??? atm

Part D

To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 1.1-L bulb, then filled it with the gas at 1.20 atm and 24.0  C and weighed it again. The difference in mass was 1.5 g . Identify the gas.

Express your answer as a chemical formula.

Identify the gas =  chemical formula ???

Explanation / Answer

Pat A. First findout no. of moles of each gases, = Wt/molecular weight

for CH4, 8 g/16.04 g/mol = 0.5 mol

for Ethane 18g / 30.07g/mol = 0.6 mol

Now we calculate partial pressure P of each gas, at 23oC (296K)

P (CH4)= nRT/ V = (0.5)(0.08206)(296)/ 10 = 1.21 atm

P(C2H6) = nRT/V = (0.6)(0.08206)(296)/ 10 = 1.46 atm

According to the law of partial presure, P (tot) = P(CH4)+ P(C2H6)+ P(C3H8)

5 = 1.21+ 1.46 +P(C3H8)

then P(C3H8) ,  5 - (1.21+ 1.46) = 2.33 atm

then Partial presure of each gas are given in order P(CH4), P(C2H6), P(C3H8) a 1.21atm, 1.46 atm and 2.33 atm.

Part B. The partial presure of O2P(O2), can be found out by the following eqn,

P(O2)= P (tot)X(O2) Where X(O2) is the mole fraction of O2

Now no.of moles of O2 = (100-36.8)/ 32= 1.975 mol

no.of moles of N2= 36.8/28.02 = 1.31 mol

total no. of moles = 1.975+ 1.31 = 3.29 mol

mol fraction of O2 = 1.975/3.29 = 0.6

Then P (O2) = (685)(0.6) = 411 mm Hg

Part C. Pressure of the gas can be measured using manometer as,

Pressure of the gas = atmospheric pressure in cm + height of Hg in cm

= (.95 atm)(76cm Hg) + 18cm = 90.2 cm Hg on converting to atm,

= 1.19 atm

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