Answer:
Explanation:
We shall find out volume of air at NTP or at 273 K and 10⁵ Pa ( 1 atm )
Let it be V₂


V₂ = 7.87 litres
22.4 litres of any gas is equivalent to 1 mole
7.87 litres of air will be equivalent to
7.87 / 22.4 moles
= .35 moles .
Answer:
x = 4.5
4.5 moles of C2H6 is required to produce 13.5 moles of H2O
Explanation:
Given equation is :
2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
if it requires 2 moles of C2H6 to produce 6 moles of H2O
then x moles of C2H6 is required to produce 13.5 moles of H2O
Using unitary method for solving, we get
x = 13.5 x 2 / 6
x = 13.5 / 3
x = 4.5
Therefore, 4.5 moles of C2H6 is required to produce 13.5 moles of H2O
Its running pressure in torr would be :
5.10 x 10^8 pa x 1kPa/1000 Pa x 760 / 101.325 kPa
= 5.10 x 10 ^5 x 76 x 10^-1 x 1.01325 x 10 ^-2
= 392.7375 x 10^2
= 3.927275 x 10^4
ope this helps