PLEASE HELP I AM STRUGGLING TO ANSWER THIS QUIZ QUESTION DUE TODAY IN 32 MINS An experiment generates 15.31 g of methane gas which occupies 17.9 L. When the experiment was conducted a second time under the same conditions. 2.15 mol of gas were created . What volume of gas gas (in L) was generated during the second trial. Give ur answer to 2 decimals. Provide answer in full calculations.

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

може ли да зареждате повече информация, моля

idk what is it? is this a question that you know the answer or u dont