Question

Butane undergoes combustion when it reacts with oxygen to produce carbon dioxide and water. 2C4H10(g)+13O2(g)⟶Δ8CO2(g)+10H2O(g)

Part A What volume, in liters, of oxygen is needed to react with 54.9 g of butane at 0.84 atm atm and 29 ∘C? Express your answer with the appropriate units.

Answer 1

Molar mass of C4H1O,

MM = 4*MM(C) + 1*MM(H) + 1*MM(O)

= 4*12.01 + 1*1.008 + 1*16.0

= 65.048 g/mol

mass(C4H1O)= 54.9 g

use:

number of mol of C4H1O,

n = mass of C4H1O/molar mass of C4H1O

=(54.9 g)/(65.05 g/mol)

= 0.844 mol

From reaction,

mol of O2 = (13/2)*number of moles of C4H10 recated

= (13/2)*0.844 mol

= 5.486 mol

Given:

P = 0.84 atm

n = 5.486 mol

T = 29.0 oC

= (29.0+273) K

= 302 K

use:

P * V = n*R*T

0.84 atm * V = 5.486 mol* 0.08206 atm.L/mol.K * 302 K

V = 162 L

Answer: 162 L