If you have 1.2754x1020 molecules of Barium nitrate (Ba(NO3)2), how many moles of barium nitrate do you have? ??? moles Ba(NO3)2 (round to 5 sig figs)

0.00021186 moles of barium nitrate

Explanation:

The given parameter is as follows;

The number of molecules of Barium nitrate Ba(NO₃)₂ = 1.2754 × 10²⁰ molecules

The number of molecule present per mole of Barium nitrate, is given by Avogadro's number, 6.02 × 10²³

1 mole of a substance contains 6.02 × 10²³ molecules

The umber of moles in a substance = (The number of molecules of the substance)/(The number of molecules per mole)

Therefore the number of moles of barium nitrate in 1.2754 × 10²⁰ molecules of barium nitrate = (1.2754 × 10²⁰)/(6.02 × 10²³) ≈ 0.00021186 moles (rounding to 5 significant figures)

The number of moles of barium nitrate present in 1.2754 × 10²⁰ molecules of barium nitrate ≈ 0.00021186 moles

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