Question

Balance the equation and then determine the liters of Oxygen need for the complete combustion of 768g of benzene (C6H6) if the reaction is run at 23.5oC and 765mHg? 2 C6H6(l) + 15 O2(g) ⟶ 12 CO2(g) + 6 H2O(g)

grams of products (3 sig figs no units):

Answer 1

Molar mass of C6H6,

MM = 6*MM(C) + 6*MM(H)

= 6*12.01 + 6*1.008

= 78.108 g/mol

mass of C6H6 = 7.68*10^2 g

mol of C6H6 = (mass)/(molar mass)

= 7.68*10^2/78.11

= 9.833 mol

Balanced chemical equation is:

2 C6H6(l) + 15 O2(g) ⟶ 12 CO2(g) + 6 H2O(g)

According to balanced equation

mol of O2 required = (15/2)* moles of C6H6

= (15/2)*9.833

= 73.74 mol

Given:

P = 765 mm Hg

= (765/760) atm

= 1.0066 atm

n = 73.74 mol

T = 23.5 oC

= (23.5+273) K

= 296.5 K

use:

P * V = n*R*T

1.0066 atm * V = 73.74 mol* 0.08206 atm.L/mol.K * 296.5 K

V = 1.78*10^3 L

Answer: 1.78*10^3 L