Respuesta :
the balanced equation for the combustion of C₆H₁₄ is as follows
2C₆H₁₄ + 19O₂ ---> 12CO₂ + 14H₂O
stoichiometry of C₆H₁₄ to O₂ is 2:19
number of moles = mass present / molar mass
number of C₆H₁₄ moles reacted = 86.17 g / 86 g/mol = 1.00 mol
according to molar ratio of 2:19
2 mol of C₆H₁₄ reacts with 19 mol of O₂
then 1.00 mol of C₆H₁₄ reacts with - 19 / 2 x 1.00 = 9.5 mol of O₂
mass of O₂ required = 9.5 mol x 32 g/mol = 304 g
mass of O₂ required is 304 g
2C₆H₁₄ + 19O₂ ---> 12CO₂ + 14H₂O
stoichiometry of C₆H₁₄ to O₂ is 2:19
number of moles = mass present / molar mass
number of C₆H₁₄ moles reacted = 86.17 g / 86 g/mol = 1.00 mol
according to molar ratio of 2:19
2 mol of C₆H₁₄ reacts with 19 mol of O₂
then 1.00 mol of C₆H₁₄ reacts with - 19 / 2 x 1.00 = 9.5 mol of O₂
mass of O₂ required = 9.5 mol x 32 g/mol = 304 g
mass of O₂ required is 304 g
304 grams mass of oxygen (O₂) is required to react completely with 86.17 grams of C₆H₁₄.
How we calculate mass from moles?
Mass of any substance will be calculated from the moles will be calculated as:
n = W/M, where
W = required mass
M = molar mass
given chemical reaction in balanced form will be written as:
2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O
First we calculate the moles of C₆H₁₄ by using the above formula as:
Moles of C₆H₁₄ = 86.17g / 86 g/mol = 1.00 mol
From the stoichiometry of the reaction, it is clear that:
2 moles of C₆H₁₄ = react with 19 moles of O₂
1 mole of C₆H₁₄ = react with 19/2 × 1 = 9.5 moles of O₂
So, mass of oxygen will be calculated by using the above formula as:
W = n × M
W = 9.5 moles × 32 g/mol = 304 g
Hence, the required mass of oxygen is 304 grams.
To know more about moles, visit the below link:
https://brainly.com/question/1034638
