2li +__h₂0= __lioh+h₂
balance the equation

2Li+2H2O=2LiOH+H2

We need to make number of the atoms of each reactant(the elements from the left side)equaled to the products(the elements from the right side)

Li is 2 on the left side so in order to make it balanced we add a 2 in front of LiOH.

For the H2O we take a look at the right side there are 2 atoms from H2 and 2 atoms of H for 2LiOH therefore we have 4 atoms of H in the right side and only 2 in the left side.So,to make them equal we will add a 2 in front of H2O.

Hope,I explained well.

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%