Question

# Carbon disulfide is an extremely flammable liquid. It can be ignited by any small spark or...

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?

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

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