Given:
K = 0.71 = Kp
The reaction of sulphur with oxygen is
S(s) + O2(g) ---> SO2(g)
initial Pressure 6.90 0
Change -x +x
Equilibrium 6.90-x x
Kp = pSO2 / pO2 = 0.71 = x / (6.90-x)
4.899 - 0.71x = x
4.899 = 1.71x
x = 2.86 atm = pressure of SO2 formed
temperature = 950 C = 950 + 273.15 K = 1223.15 K
Volume = 50 L
Let us calculate moles of SO2 formed using ideal gas equation as
PV = nRT
R = gas constant = 0.0821 L atm / mol K
putting other values
n = PV / RT = 2.86 X 50 / 1223.15 X 0.0821 = 1.42 moles
Moles of Sulphur required = 1.42 moles
Mass of sulphur required or consumed = moles X atomic mass of sulphur
mass of S = 1.42 X 32 = 45.57 grams or 0.04557 Kg of sulphur
Answer:
the company has also fished
Answer:
11 1/2 cm I think lol
Explanation:
it is in between the 11 and 12 mark
