The decomposition of so3 to so2 gas and o2 gas can be described in the balanced chemical equation:
2so3(g) = 2 so2 (g) + 02(g).
so assuming a complete reaction, the ratio of so2 gas to total products is 2/3 while that of 02 is 1/3.
Subtracting water's water vapor pressure, 760-40 mm hG = 720 mm Hg.
then the products partial pressures are
so2 = 2/3 * (720) = 480 mm Hg.
o2 = 720-480 = 240 mm Hg.
To Find :
The volume of 12.1 moles hydrogen at STP.
Solution :
We know at STP, 1 mole of gas any gas occupy a volume of 22.4 L.
Let, volume of 12.1 moles of hydrogen is x.
So, x = 22.4 × 12.1 L
x = 271.04 L
Therefore, the volume of hydrogen gas at STP is 271.04 L.