Question

Octate burns according to the following balanced chemical equation.

2C₈H₁₈ + 25O₂ -> 16CO₂ + 18H₂O

A cylinder in an automobile engine has a volume of 494mL. If the cylinder is filled with 0.210 atm of oxygen at 50 celcius, what mass (in grams) of octane must be injected to react with all of the oxygen present?

A) 0.0427g

B) 0.231g

C) 0.0724g

D) 0.0358g

D is the correct answer. I tried using PV = (mass/molar mass) RT, but I had the stoichiometry part of it wrong. Please explain your answer. Thanks!

Answer 1

we have:

P = 0.21 atm

V = 494.0 mL

= (494.0/1000) L

= 0.494 L

T = 50.0 oC

= (50.0+273) K

= 323 K

find number of moles using:

P * V = n*R*T

0.21 atm * 0.494 L = n * 0.08206 atm.L/mol.K * 323 K

n = 3.912*10^-3 mol

This is number of mol of O2 reacting

from given equation,

moles of C8H18 = (2/25)*moles of O2

= (2/25)*3.912*10^-3 mol

= 3.13*10^-4 mol

Molar mass of C8H18 = 8*MM(C) + 18*MM(H)

= 8*12.01 + 18*1.008

= 114.224 g/mol

we have below equation to be used:

mass of C8H18,

m = number of mol * molar mass

= 3.13*10^-4 mol * 114.224 g/mol

= 0.0358 g

Answer: 0.0358 g