Question

(a) How many grams of CO2 (44.01 g/mol) can be produced by 75.0 g of octane (114.22 g/mol) reacting with sufficient oxygen?

(b) How many grams of O2 (32.00 g/mol) are required for this reaction?

2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(g)

Answer 1

a)


mass of C8H18 = 75 g
molar mass of C8H18 = 114.22 g/mol
mol of C8H18 = (mass)/(molar mass)
= 75/114.22
= 0.6566 mol


Balanced chemical equation is:
2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(g)


According to balanced equation
mol of CO2 formed = (16/2)* moles of C8H18
= (16/2)*0.6566
= 5.253 mol



mass of CO2 = number of mol * molar mass
= 5.253*44.01
= 231.2 g
Answer: 231.2 g

b)
According to balanced equation
mol of O2 formed = (25/2)* moles of C8H18
= (25/2)*0.6566
= 8.208 mol


mass of O2 = number of mol * molar mass
= 8.208*32.00
= 262.6 g
Answer: 262.6 g