I might go with d it seems right to me hope this helps.
* Mole ratio:
C2H4 + 3 O2= 2 CO2 + 2 H2O
1 mole C2H4 --------------- 2 moles H2O
8.00 moles C2H4 ---------- ?
8.00 x 2 / 1 => 16 moles of H2O
Therefore:
1 mole --------- 22.4 L at ( STP)
16 moles ------- ?
16 x 22.4 / 1 => 358.4 L
hope this helps!
Answer:
12.8 g of
must be withdrawn from tank
Explanation:
Let's assume
gas inside tank behaves ideally.
According to ideal gas equation- 
where P is pressure of
, V is volume of
, n is number of moles of
, R is gas constant and T is temperature in kelvin scale.
We can also write, 
Here V, T and R are constants.
So,
ratio will also be constant before and after removal of
from tank
Hence, 
Here,
and 
So, 
So, moles of
must be withdrawn = (0.66 - 0.26) mol = 0.40 mol
Molar mass of
= 32 g/mol
So, mass of
must be withdrawn = 
Conduction should be the correct answer.