Consider two gas cylinders of the same volume and temperature, one containing 1.0 mol of propane, c3h8,c3h8, and the other 2.0 mol of methane, ch4.ch4. which of the following statements is true?
a. the c3h8c3h8 and ch4ch4 molecules have the same urmsurms
b. the c3h8c3h8 and ch4ch4 molecules have the same average kinetic energy
c. the rate at which the molecules collide with the cylinder walls is the same for both cylinders
d. the gas pressure is the same in both cylinders

B

Explanation:

Statement A

Vrms (root-mean-square velocity) is defined as follows.

Vrms = √(2RT/Mm), where R is the ideal gas constant, T is the temperature, and Mm is the molar mass of the gas. R and T are the same for both gases, but the molar mass is different, so Vrms for the two gases is different

Statement B

Kinetic energy KE of a molecule is related to temperature as follows:

KE = 3/2KbT, where Kb is the Boltzmann's constant and T is the temperature. The temperature is the same for both gases, so KE is the same for both.

Statement C

The rate depends on the number of molecules present. Since there are twice as many molecules of methane as propane, the rate is different for both gases.

Statement D

Pressure is directly proportional to the number of moles, based on the ideal gas law (pV = nRT). The volume and temperature are the same for both gases, so since there are twice as many molecules of methane as propane, the pressure is twice as great in the cylinder of methane.

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g