If 152 grams of ethane (c2h6) are reacted with 231 grams of oxygen gas, what is the mass of the excess reactant leftover after the reaction has reached completion? 2c2h6(g) + 7o2(g) → 4co2(g) + 6h2o(g)

The answer is:

the mass of the excess reactant  leftover after the reaction has reached completion is 90.135 grams

The explanation:

According to the reaction equation:

 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g)

when m is the mass of C₂H₆ m(C₂H₆) = 152 g

So we need to get the number of moles of C₂H₆

n(C₂H₆) = mass C₂H₆ / molar mass of C₂H₆(M)

and when the molar mass of C₂H₆ = 30 g/mol

so, by substitution:

n(C₂H₆) = m(C₂H₆) / M(C₂H₆).

n(C₂H₆) = 152 g / 30 g/mol.

n(C₂H₆) = 5.067 mol.

Then

when the mass of O₂ m(O₂) = 231 g

so we need to get the number of moles of O₂

when nO₂ = mass O₂/ molar mass of O₂

when molar mass of O₂ = 32 g /mol

So, by substitution:

n(O₂) = 231 g / 32 g/mol.

n(O₂) = 7.218 mol

So O₂ is the limiting reactant

according to the chemical reaction we can get the molar ratio between the O₂and C₂H₆:

n O₂ : n C₂H₆    →  7.218 : n C₂H₆

      7 : 2                   7      :   2

∴ n(C₂H₆) = 2 * 7.218 mol / 7

∴ n(C₂H₆) = 2.0625 mol

The number of moles remaining n(C₂H₆) = 5.067 mol - 2.0625 mol

∴ n (C₂H₆) = 3.0045 mol

So the mass remains = moles remains * molar mass of C₂H₆

∴ m (C₂H₆) = 3.0045 mol * 30 g/mol = 90.135 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

passing a magnet thought the

mixture , because iron gets attracted to magnet .