Using the equation below, what is the correct form of the conversion factor needed to convert the number of moles of h2o to the number of moles of nh3 produced?

mg3n2 + 6h2o --> 3mg(oh)2 + 2nh3

2 mol nh3
6 mol h2o

18 g h2o
17 g nh3

1 mol mg3n3
6 mol h2o

18 g h2o
1 mol h2o

The conversion factor would be 2 mol NH3 to 6 mol H20

From the balanced equation of the reaction below:

The mole ratio of H2O to NH3 is 6:2. In other words, for every 6 moles of H2O, 2 moles of NH3 would be produced.

This also means that for every 2 moles of NH3 produced, 6 moles of water would have been consumed by the reaction.

More on mole ratios can be found here: link

2 mol NH3

6 mol H2O

Explanation: from the equation, 6 moles of H2O will produce 2 moles of NH3

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

it is the most commonly accepted one