4 moles as their raport is 4/2 or 2/1
setup 1 : to the right
setup 2 : equilibrium
setup 3 : to the left
<h3>Further explanation</h3>
The reaction quotient (Q) : determine a reaction has reached equilibrium
For reaction :
aA+bB⇔cC+dD
![\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Comparing Q with K( the equilibrium constant) :
K is the product of ions in an equilibrium saturated state
Q is the product of the ion ions from the reacting substance
Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)
Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium
Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)
Keq = 6.16 x 10⁻³
Q for reaction N₂O₄(0) ⇒ 2NO₂(g)
![\tt Q=\dfrac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
Setup 1 :

Q<K⇒The reaction moved to the right (products)
Setup 2 :

Q=K⇒the system at equilibrium
Setup 3 :

Q>K⇒The reaction moved to the left (reactants)
<span>2 C6H6 + 15 O2 = 12 CO2 + 6 H2O
</span>
2 moles C6H6 -------------------- 6 moles H2O
8.73 moles C6H6 --------------- ( moles H2O)
( moles H2O) = 8.73 x 6 / 2
( moles H2O) = 52.38 / 2
= 26.19 moles of H2O
Molar mass H2O = 18.00 g/mol
1 mole -------------- 18.00 g
26.19 moles ----- ( mass of H2O)
mass of H2O = 26. 19 x 18.00 => 471.42 g of water
hope this helps!