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)
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
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