Question

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)

Answer 1

Answer is: the mass of the excess reactant (ethane) leftover is 90.135 grams.
Chemical reaction: 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g).
m(C₂H₆) = 152 g.
n(C₂H₆) = m(C₂H₆) ÷ M(C₂H₆).
n(C₂H₆) = 152 g ÷ 30 g/mol.
n(C₂H₆) = 5.067 mol.
m(O₂) = 231 g.
n(O₂) = 231 g ÷ 32 g/mol.
n(O₂) = 7.218 mol; limiting reactant.
From chemical reaction: n(O₂) : n(C₂H₆) = 7 : 2.
n(C₂H₆) = 2 · 7.218 mol ÷ 7.
n(C₂H₆) = 2.0625mol.
Δn(C₂H₆) = 5.067 mol - 2.0625 mol.
Δn(C₂H₆) = 3.0045 mol.
Δm(C₂H₆) = 3.0045 mol · 30 g/mol = 90.135 g.

Answer 2

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