Question

If 5.0 g S8 is made to react with 2.0 L O2 at 30 oC and 1.0 atm, what is the maximum amount (in g) of SO3 produced? ½ S8 (s) + 6 O2 (g)  4 SO3 (g)

Answer 1

Number of moles of S8 = 5.0 g / 256 g/mol = 0.0195 mole

PV = nRT

where, P = pressure = 1.0 atm

V = volume = 2.0 L

n = number of moles

R = Gas constant

T = temperature = 30 + 273 = 303 K

1.0 * 2.0 = n * 0.0821 * 303

2.0 = n * 24.9

n = 2.0 / 24.9 = 0.0803 mole

Therefore, the number of moles of O2 = 0.0803 mole

From the balanced equation we can say that

0.5 mole of S8 requires 6 mole of O2 so

0.0195 mole of S8 will require

= 0.0195 mole of S8 *( 6 mole of O2 / 0.5 mole of S8)

= 0.234 mole of O2

But we have 0.0803 mole of O2 which is in short so O2 is limiting reactant

From the balanced equation we can say that

6 mole of O2 produces 4 mole of SO3 so

0.0803 mole of O2 will produce

= 0.0803 mole of O2 *(4 mole of SO3 / 6 mole of O2)

= 0.0535 mole of SO3

mass of 1 mole of SO3 = 80.066 g so

the mass of 0.0535 mole of SO3 = 4.28 g

Therefore, the mass of SO3 produced would be 4.28 g