Question

What mass of carbon dioxide (co2) can be produced from 86.17 grams of c6h14 and excess oxygen?

Answer 1

From 86.17 grams of C₆H₁₄ and excess oxygen, approximately 264.054 grams of CO₂ can be produced by first balancing the combustion equation, then using stoichiometry to find the moles of C₆H₁₄, and finally converting those moles to mass of CO₂.

Step-by-step explanation:

The question asks for the mass of carbon dioxide (CO₂) that can be produced from 86.17 grams of C₆H₁₄ (hexane) when reacted with excess oxygen. To answer this, we'll require the balanced chemical equation for the combustion of hexane and then use stoichiometry to calculate the mass of CO₂ produced.

Firstly, let's balance the chemical equation for the combustion of hexane:

2 C₆H₁₄(l) + 19 O₂(g) → 12 CO₂(g) + 14 H₂O(g)

Next, we'll calculate the moles of C₆H₁₄ using its molar mass (86.18 g/mol for C₆H₁₄):

Moles of C₆H₁₄ = Mass / Molar mass = 86.17 g / 86.18 g/mol = ~1 mol C₆H₁₄

According to the balanced equation, the combustion of 2 moles of C₆H₁₄ produces 12 moles of CO₂. Since we have approximately 1 mole of C₆H₁₄, it will produce approximately 6 moles of CO₂.

Finally, we convert the moles of CO₂ to mass:

Mass of CO₂ = Moles of CO₂ x Molar mass of CO₂ = 6 moles x 44.009 g/mol = 264.054 g CO₂

So, 86.17 grams of C₆H₁₄ with excess oxygen will produce approximately 264.054 grams of CO₂.

Answer 2

2C6H14 + 13O2 ---> 6CO2 +14H2O

M(C6H14)=12.011*6 +1.008*14 ≈ 86.17 g/mol

86.17 g C6H14 is 1 mole.

2C6H14 + 13O2 ---> 6CO2 +14H2O
from reaction 2 mol 6 mol
from the problem 1 mol 3 mol

M(CO2)= 12.011 + 2*15.999= 44.009 g/mol
3 mol CO2*44.009 g/1 mol CO2 ≈ 132.0 g CO2
Answer : 132.0 g CO2