What means of no2 is contained in a 13.0l tank at 464 kpa and 112 celsius 289g 191 g 86.5 g 1.88 g

= 86.5 g

Explanation:

Using the equation;

PV = n RT

where P is the pressure, 464 kpa or 4.58 atm

V is the volume , 13.0 l

n is the number of moles

R is the ideal gas constant , 0.0821 atm L/mol/K

T is the temperature, 112 + 273.15 = 385.15 K

Therefore;

n = PV/RT

n = (4.58 × 13)/(0.0821×385.15)

  = 1.883 moles

The mass will be;

 = 1.883 moles × 46.0 g/mol

 = 86.57 g

 = 86.5 g

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

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