Question

Carbon disulfide is an extremely flammable liquid. It can be ignited by any small spark or even a very hot surface such as a steam pipe. The combustion reaction is CS2 (g) + 3O2(g) → CO2 (g) + 2SO2(g)

When 11.0 g of CS2 are burned in excess oxygen, how many liters of CO2 and SO2 are formed at 28 °C and 883 torr?

Answer 1

Molar mass of CS2,

MM = 1*MM(C) + 2*MM(S)

= 1*12.01 + 2*32.07

= 76.15 g/mol

mass(CS2)= 11.0 g

number of mol of CS2,

n = mass of CS2/molar mass of CS2

=(11.0 g)/(76.15 g/mol)

= 0.1445 mol

Balanced chemical equation is:

CS2 + 3 O2 ---> CO2 + 2 SO2

1)

According to balanced equation

mol of CO2 formed = (1/1)* moles of CS2

= (1/1)*0.1445

= 0.1445 mol

we have

P = 883.0 torr

= (883.0/760) atm

= 1.1618 atm

n = 0.1445 mol

T = 28.0 oC

= (28.0+273) K

= 301 K

use:

P * V = n*R*T

1.1618 atm * V = 0.1445 mol* 0.0821 atm.L/mol.K * 301 K

V = 3.07 L

Answer: volume of CO2 = 3.07 L

2)

According to balanced equation

mol of SO2 formed = (2/1)* moles of CS2

= (2/1)*0.1445

= 0.2889 mol

n = 0.2889 mol

use:

P * V = n*R*T

1.1618 atm * V = 0.2889 mol* 0.0821 atm.L/mol.K * 301 K

V = 6.14 L

Answer: volume of SO2 = 6.14 L