Answer:
77.14 atm of pressure should be of an acetylene in the tank.
Explanation:
According to reaction, 2 moles of acetylene reacts with 5 moles of oxygen.
Moles of oxygen=
Moles of acetylene =
Volume of large tank with oxygen gas,
Pressure of the oxygen gas inside the tank =
..[1]
Volume of small tank with acetylene gas ,
Pressure of the acetylene gas inside the tank =
..[2]
Considering both the gases having same temperature T, [1]=[2]
77.14 atm of pressure should be of an acetylene in the tank.
Answer:
Explanation:
We shall find volume of gas at NTP or at 273 K , 760 mm of Hg .
Pressure of given gas = 1.06 x 760 mm of Hg less vapor pressure of water .
= 805.6 - 23.76 = 781.84 mm of Hg
For it we use gas law formula ,
P₁V₁ / T₁ = P₂V₂ / T₂
781.84 x 136.1 / ( 273 + 25 ) = 760 x V₂ / 273
= 128.26 mL .
= 128.26 x 10⁻³ L .
22.4 L of oxygen will have mass of 32 g
128.26 x 10⁻³ L of oxygen will have mass of 32 x 128.26 x 10⁻³ / 22.4 g
= 183.22 mg .
At STP (standard temperature and pressure) 1 mole of a gas takes up 22.4L
so 2.33/22.4=0.104 mol H2S