:a ne gas sample has a pressure of 1.20 atm and volume of 35.6 l at 27.0ºc. what is the sample volume if the pressure changes to 3.60 atm and the temperature to 77.0ºc?

13.84 L of Ne gas

Explanation:

The exercise can be resolved by applying the General Gas Law, which establishes the relationship between the Pressure, Volume and Temperature of a gas in its initial state and its final state. Mathematically, it is:

To use this equation, you need to convert the temperatures from Celsius scale to Kelvin scale (absolute scale), and to make this conversion the formula is used:

T (K) = t (° C) + 273.15

We replace the values and obtain the absolute temperatures of the initial state (T₁) and the final state (T₂)

T₁ (K) = 27.0 ° C + 273.15 = 300.15 K

T₂ (k) = 77.0 ° C + 273.15 = 350.15 K

Clearing in the equation of the general gas law V₂, and replacing the other values provided by the statement, the equation and result would be:

Therefore, the final volume occupied by the gas sample Ne is 13.84 L

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

58750.08068 is the answer i think